Secure active element machine

ABSTRACT

Based upon the principle of Turing incomputability, and novel properties of the Active Element Machine, a malware-resistant computing machine is constructed. This new computing machine is a non-Turing, non-register machine (non von-Neumann), called an Active Element Machine (AEM). AEM programs are designed so that the purpose of the computation is difficult to apprehend by an adversary and hijack with malware. These methods can help hinder reverse engineering of proprietary algorithms and hardware design. 
     Using quantum randomness, the AEM can deterministically execute a universal digital computer program with active element firing patterns that are Turing incomputable. In some embodiments, a more powerful computational procedure is demonstrated than Turing&#39;s computational procedure (digital computer procedure). Current digital computer algorithms can be derived or designed with a Turing machine computational procedure. A novel class of computing machines is built where the purpose of the program&#39;s execution is difficult to apprehend (Turing incomputable).

RELATED APPLICATIONS

This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/462,260, entitled “Navajo Active Element Machine” filed Jan. 31, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/465,084, entitled “Unhackable Active Element Machine” filed Mar. 14, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/571,822, entitled “Unhackable Active Element Machine Using Randomness” filed Jul. 6, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/572,607, entitled “Unhackable Active Element Machine Unpredictable Firing Interpretations” filed Jul. 18, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/572,996, entitled “Unhackable Active Element Machine with Random Firing Interpretations and Level Sets” filed Jul. 26, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/626,703, entitled “Unhackable Active Element Machine with Turing Undecidable Firing Interpretations” filed Sep. 30, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. 61/628,332, entitled “Unhackable Active Element Machine with Turing Incomputable Firing Interpretations” filed Oct. 28, 2011, which is incorporated herein by reference. This application claims priority benefit of U.S. Provisional Patent Application Ser. No. ______, entitled “Unhackable Active Element Machine with Turing Incomputable Computation” filed Nov. 7, 2011, which is incorporated herein by reference.

FIELD

The specification generally relates to computing.

BRIEF DESCRIPTION OF FIGURES AND TABLES

In the following figures and tables, although they may depict various examples of the invention, the invention is not limited to the examples depicted in the figures and tables.

FIG. 1 shows a fire pattern of 0000 for active elements X₀, X₁, X₂, and X₃. Elements X₀, X₁, X₂, and X₃ don't fire during window W.

FIG. 2 shows a fire pattern of 0001 for active elements X₀, X₁, X₂, and X₃. During window W, elements X₀, X₁, X₂ don't fire and X₃ fires.

FIG. 3 shows a fire pattern of 0010 for active elements X₀, X₁, X₂, and X₃. During window W, only element X₂ fires.

FIG. 4 shows a fire pattern of 0011 for active elements X₀, X₁, X₂, and X₃. During window W, only elements X₂ and X₃ fire.

FIG. 5 shows a fire pattern of 0100 for active elements X₀, X₁, X₂, and X₃. During window W, only element X₁ fires.

FIG. 6 shows a fire pattern of 0010 for active elements X₀, X₁, X₂, and X₃ used to compute 1⊕0.

FIG. 7 shows a fire pattern of 0010 for active elements X₀, X₁, X₂, and X₃ used to compute 0⊕1.

FIG. 8 shows a fire pattern of 1011 for active elements X₀, X₁, X₂, and X₃ used to compute 1⊕1.

FIG. 9 shows a fire pattern of 0100 for active elements X₀, X₁, X₂, and X₃ used to compute 0⊕0.

FIG. 10 shows a fire pattern of 0101 for active elements X₀, X₁, X₂, and X₃ used to compute

(0

0)=1.

FIG. 11 shows a fire pattern of 0101 for active elements X₀, X₁, X₂, and X₃ used to compute

(1

0)=1.

FIG. 12 shows a fire pattern of 0101 for active elements X₀, X₁, X₂, and X₃ used to compute

(0

1)=1.

FIG. 13 shows a fire pattern of 1010 for active elements X₀, X₁, X₂, and X₃ used to compute

(1

1)=0.

FIG. 14 shows the geometry of a vertical cover.

FIG. 15 shows the geometry of a horizontal cover.

FIG. 16 shows the geometry when the map is not a vertical cover.

FIG. 17 shows the geometry of a vertical cover for the Vertical Covering Fixed Point Theorem in 1.13.

FIG. 18 shows the geometry of a horizontal cover for the Horizontal Covering Fixed Point Theorem in 1.14.

FIG. 19 shows a Turing machine configuration (q, k, T) with the machine in state q, tape head at tape square k and containing alphabet symbol T(k) in tape square k.

FIG. 20 shows the window of execution for one cycle of a periodic point p=[q, 12

1

212222].

FIG. 21 shows the pertinent parts of the machine configuration used to determine the unit square domain of a left affine map (2.16) or right affine map (2.19).

FIG. 22 shows a Turing machine computational step that corresponds to one iteration of a corresponding left affine function (2.16).

FIG. 23 shows a Turing machine computational step that corresponds to one iteration of a right affine function (2.19)

FIG. 24 shows case A of the definition the Edge Pattern Substitution Operator.

FIG. 25 shows case B of the definition the Edge Pattern Substitution Operator.

FIG. 26 shows case C of the definition the Edge Pattern Substitution Operator.

FIG. 27 shows case D of the definition the Edge Pattern Substitution Operator.

FIG. 28 shows 8 points of the affine map orbit corresponding to the aperiodic machine defined in 6.2.

FIG. 29 shows 4 additional points of the affine map orbit in 6.2.

FIG. 30 shows 9 points of the affine map orbit corresponding to the periodic machine defined in 6.3.

FIG. 31 shows 9 additional points of the affine map orbit corresponding to the periodic machine defined in 6.3.

FIG. 32 shows 7 additional points of the affine map orbit corresponding to the periodic machine defined in 6.3.

FIG. 33 shows a square about point z_(n) referred to in 6.20.

FIG. 34 shows a vertical cover referred to in 6.20.

For each boolean function ƒ: {0, 1}×{0, 1}→{0, 1}, table 1 shows the level set rules used to design an active element machine program that separate elements of {(0, 0), (1, 0), (0, 1), (1, 1)}.

Table 2 shows Minsky's universal Turing machine.

Table 3 shows the Boolean version of the universal Turing machine in table 2.

Table 4 shows the level set rules for η₀: the 0^(th) bit of Turing program η in table 3.

Table 5 shows the level set rules for η₁: the 1^(st) bit of Turing program η in table 3.

Table 6 shows the level set rules for η₂: the 2^(nd) bit of Turing program η in table 3.

Table 7 shows the level set rules for η₃: the 3^(rd) bit of Turing program η in table 3.

Table 8 shows the level set rules for η₄: the 4^(th) bit of Turing program η in table 3.

Table 9 shows the level set rules for η₅: the 5^(th) bit of Turing program η in table 3.

Table 10 shows all sixteen firing patterns of elements X₀, X₁, X₂, X₃ which represents four bits that are added using elements C₀, C₁, C₂, C₃ and elements P₀, P₁, P₂, P₃.

Table 11 shows the amplitudes from elements X₀, X₁, X₂, X₃ to elements C₀, C₁, C₂, C₃ and thresholds for elements C₀, C₁, C₂, C₃. Table 12 shows the amplitudes from elements C₀, C₁, C₂, C₃ to elements P₀, P₁, P₂, P₃ and thresholds for elements P₀, P₁, P₂, P₃.

Table 13 shows four bit multiplication where one four bit number is y₃ y₂ y₁ y₀ and the other four bit number is z₃ z₂ z₁ z₀ and the result is e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀

Table 14 shows the amplitude and threshold used to compute the value of e₀. Table 15 shows the firing patterns for elements S₁₀ and S₀₁ representing the value of products y₀ z₀ and y₀ z₁. Table 16 shows the amplitudes from elements S₁₀ and S_(0l) to elements C_(0l) and C₁₁ and the thresholds of C₀₁ and C₁₁. Table 17 shows the amplitude and threshold used to compute the value of e₁.

Table 18 shows the firing patterns for elements S₂₀, S₁₁, S₀₂ and C₁₁. Table 19 shows the amplitudes from elements S₂₀, S₁₁ S₀₂, C₁₁ to elements C₀₂, C₁₂, C₂₂ C₃₂ and thresholds of C₀₂, C₁₂, C₂₂ and C₃₂. Table 20 shows the amplitudes from elements C₀₂, C₁₂, C₂₂, C₃₂ to elements P₀₂, P₁₂, P₂₂ and the thresholds of elements P₀₂, P₁₂, P₂₂. Table 21 shows the amplitude and threshold used to compute the value of e₂.

Table 22 shows the firing patterns for elements S₃₀, S₂₁, S₁₂, S₀₃, P₁₂ representing the value of products y₃ z₀, y₂ z₁, y₁ z₂ and y₀ z₃ and the carry value. Table 23 shows the amplitudes from elements S₃₀, S₂₁, S₁₂ and S₀₃ to elements C₀₃, C₁₃, C₂₃, C₃₃, and C₄₃. Table 24 shows the amplitudes from elements C₀₃, C₁₃, C₂₃, C₃₃, and C₄₃ to elements P₀₃, P₁₃, P₂₃ and the thresholds of elements P₀₃, P₁₃, P₂₃. Table 25 shows the amplitude and threshold used to compute the value of e₃.

Table 26 shows the firing patterns for elements S₃₁, S₂₂, S₁₃, P₁₃, P₂₂. Table 27 shows the amplitudes from elements S₃₁, S₂₂, S₁₃, P₁₃, P₂₂ to elements C₀₄, C₁₄, C₂₄, C₃₄, C₄₄ and the thresholds of C₀₄, C₁₄, C₂₄, C₃₄ and C₄₄.

Table 28 shows the amplitudes from elements C₀₄, C₁₄, C₂₄, C₃₄, and C₄₄ to elements P₀₄, P₁₄, P₂₄ and the thresholds of elements P₀₄, P₁₄, P₂₄. Table 29 shows the amplitude and threshold used to compute the value of e₄.

Table 30 shows the firing patterns for elements S₃₂, S₂₃, P₁₄, P₂₃. Table 31 shows the amplitudes from elements S₃₂, S₂₃, P₁₄, P₂₃ to elements C₀₅, C₁₅, C₂₅, C₃₅ and the thresholds of C₀₅, C₁₅, C₂₅, C₃₅. Table 32 shows the amplitudes from elements C₀₅, C₁₅, C₂₅, C₃₅ to elements P₀₅, P₁₅, P₂₅ and the thresholds of elements P₀₅, P₁₅, P₂₅. Table 33 shows the amplitude and threshold used to compute the value of e₅.

Table 34 shows the firing patterns for elements S₃₃, P₁₅, P₂₄. Table 35 shows the amplitudes from elements S₃₃, P₁₅, P₂₄ to elements C₀₆, C₁₆, C₂₆ and the thresholds of C₀₆, C₁₆, C₂₆. Table 36 shows the amplitudes from elements C₀₆, C₁₆, C₂₆ to elements P₀₆, P₁₆ and the thresholds of elements P₀₆, P₁₆. Table 37 shows the amplitude and threshold used to compute the value of e₆.

Table 38 shows some details of the four bit multiplication 1110*0111. Table 39 shows some details of the four bit multiplication 1011*1001. Table 40 shows some details of the four bit multiplication 1111*1110. Table 41 shows some details of the four bit multiplication 1111*1111.

BRIEF SUMMARY OF THE INVENTION

What are you trying to do? Why is this compelling?

Based upon the principles of Turing incomputability and connectedness and novel properties of the Active Element Machine, a malware-resistant computing machine is constructed. This novel computing machine is a non-Turing, non-register machine (non von-Neumann), called an active element machine (AEM). AEM programs are designed so that the purpose of the AEM computations are difficult to apprehend by an adversary and hijack with malware. As a method of protecting intellectual property, these methods can also be used to help thwart reverse engineering of proprietary algorithms, hardware design and other areas of intellectual property.

LIMITATIONS AND DEFICIENCIES OF PRIOR ART

How is it done at present? What are the limitations of present cybersecurity approaches?

Some prior art (approaches) has tried to conceal and protect a computation by enclosing it in a physical barrier, or by using a virtual barrier, e.g. firewall, or private network. The prior art has not been successful at securing computers, networks and the Internet. Operating system weaknesses and the proliferation of mobile devices and Internet connectivity have enabled malware to circumvent these boundaries.

In regard to confidentiality of data, some prior art uses cryptography based on the P≠NP complexity assumption, which relies on large enough computing bounds to prevent breaking the cryptography. In the future, these approaches may be compromised by more advanced methods such as Shor's algorithm, executing on a quantum computing machine.

In the case of homomorphic cryptography <http://crypto.stanford.edu/craig/> its computing operations are about twelve orders of magnitude too slow and the operations execute on a register machine. Homomorphic cryptography assumes that the underlying encryption E operations obey the homomorphism ring laws E(x+y)=E(x)+E(y) and E(x)·E(y)=E(x·y)<http://tinyurl.com/4csspud>. If the encrypted execution is tampered with (changed), then this destroys the computation even though the adversary may be unable to decrypt it. This is analogous to a DDoS attack in that you don't have to be able to read confidential data to breach the cybersecurity of a system. Homomorphic cryptography executing on a register machine along with the rest of the prior art is still susceptible to fundamental register machine weaknesses discussed below.

Some prior art has used the evolution of programs executing on a register machine (von=Neumann architecture) architecture. [Fred Cohen, “Operating Systems Protection Through Program Evolution”, IFIP-TC11 ‘Computers and Security’ (1993) V12#6 (October 1993) pp. 565-584].

The von Neumann architecture is a computing model for a stored-program digital computer that uses a CPU and a separate structure (memory) to store both instructions and data. Generally, a single instruction is executed at a time in sequential order and there is no notion of time in von-Neumann machine instructions: This creates attack points for malware to exploit. Some prior art has used obfuscated code that executes on a von-Neumann architecture. See <http://www.ioccc.org/main.html> on the International Obfuscated C code contest.

In the prior art, computer program instructions are computed the same way at different instances: fixed representation of the execution of a program instruction. For example, the current microprocessors have the fixed representation of the execution of a program instruction property. (See http://en.wikipedia.org/wiki/Microprocessor.) The processors made by Intel, Qualcomm, Samsung, Texas Instrument and Motorola use a fixed representation of the execution of their program instructions. (See www.intel.com http://en.wiki edia.org/wiki/Intel_processor, http://www.qualcomm.com/, www.samsung.com and http://www.ti.com/)

The ARM architecture, which is licensed by many companies, uses a fixed representation of the execution of its program instructions. (See www.arm.com and www.wikipedia.org/wiki/Arm_instruction_set.) In the prior art, not only are the program instructions computed the same way at different instances, there are also a finite number of program instructions representable by the underlying processor architecture. This affects the compilation of a computer program written into the processor's (machine's) program instructions. As a consequence, the compiled machine instructions generated from a program written in a programming language such as—C, JAVA, C++, Fortran, assembly language, Ruby, Forth, LISP, Haskell, RISC machine instructions, java virtual machine, Python or even a Turing machine program—are computed the same way at different instances. This fixed representation of the execution of a program instruction property in the prior art makes it easier for malware to exploit security weaknesses in these computer programs.

Some prior art relies on operating systems that execute on a register machine architecture. The register machine model creates a security vulnerability because its computing steps are disconnected. This topological property (disconnected) creates a fundamental mathematical weakness in the register machine so that register machine programs may be hijacked by malware. Next, this weakness is explained from the perspective of a digital computer program (computer science).

In DARPA's CRASH program <http://tinyurl.com/4 khv28q>, they compared the number of lines of source code in security software written over twenty years versus malware written over the same period. The number of lines of code in security software grew from about 10,000 to 10 million lines; the number of lines of code in malware was almost constant at about 125 lines. It is our thesis that this insightful observation is a symptom of fundamental security weakness(es) in digital computer programs (prior art of register machines): It still takes about the same number of lines of malware code to hijack digital computer's program regardless of the program's size.

The sequential execution of single instructions in the register and von-Neumann machine make the digital computer susceptible to hijacking and sabotage. As an example, by inserting just one jmp WVCTF instruction into the program or changing the address of one legitimate jmp instruction to WVCTF, the purpose of the program can be hijacked.

Malware Instructions (polymorphic variant) WVCTF: mov eax, drl jmp Loc1 Loc2: mov edi, [eax] LOWVCTF: pop ecx jecxz SFMM inc eax mov esi, ecx dec eax nop mov eax, 0d601h jmp Loc3 Loc1: mov ebx, [eax+10h] jmp Loc2 Loc3: pop edx pop ecx nop call edi jmp LOWVCTF SFMM: pop ebx Pop eax stc

From a Turing machine (TM) perspective, only one output state r of one TM program, command η(q, a)=(r, b, x) needs to be changed to state m combined with additional hijacking TM commands adjoined to the original TM program. After visiting state m, these hijacking commands are executed, which enables the purpose of the original TM program to be hijacked.

Furthermore, once the digital computer program has been hijacked, if there is a friendly routine to check if the program is behaving properly, this safeguard routine will never get executed. As a consequence, the sequential execution of single instructions cripples the register machine program from defending and repairing itself. As an example of this fundamental security weakness of a digital computer, while some malware may have difficulty decrypting the computations of a homomorphic encryption operation, the malware can still hijack a register machine program computing homomorphic encryption operations and disable the program.

BRIEF SUMMARY OF NOVELTY AND ADVANTAGES OVER PRIOR ART

What is novel about the Secure Active Element Machine?

A. A novel non-Turing computing machine—called the active element machine—is presented that has new capabilities. Turing machine, digital computer programs, register machine programs and standard neural networks have a finite prime directed edge complexity. (See definition 4.23.) A digital computer program or register machine program can be executed by a Turing machine. (See [7], [20] and [24]).

An active element machine (AEM) that has unbounded prime directed edge complexity can be designed or programmed. This is important advantage because rules describing a AEM program are not constant as a function of time. Furthermore, these rules change unpredictably because the AEM program interpretation can be based on randomness and in some embodiments uses quantum randomness. In some embodiments, quantum randomness uses quantum optics or quantum phenomena from a semiconductor. The changing the rules property of the AEM programs with randomness makes it difficult for malware to apprehend the purpose of an AEM program.

B. Meta commands and the use of time enable the AEM to change its program as it executes, which makes the machine inherently self-modifying. In the AEM, self-modification of the connection topology and other parameters can occur during a normal run of the machine when solving computing problems. Traditional multi-element machines change their architecture only during training phases, e.g. when training neural networks or when evolving structures in genetic programming. The fact that self-modification happens during runtime is an important aspect for cybersecurity of the AEM. Constantly changing systems can be designed that are difficult to reverse engineer or to disable in an attack. When the AEM has enough redundancy and random behavior when self-modifying, multiple instances of an AEM—even if built for the same type of computing problems—all look different from the inside. As a result, machine learning capabilities are built right into the machine architecture. The self-modifying behavior also enables AEM programs to be designed that can repair themselves if they are sabotaged.

C. The inherent AEM parallelism and explicit use of time can be used to conceal the computation and greatly increase computing speed compared to the register machine. There are no sequential instructions in an AEM program. Multiple AEM commands can execute at the same time. As a result, AEM programs can be designed so that additional malware AEM commands added to the AEM program would not effect the intended behavior of the AEM program. This is part of the topological connectedness.

D. An infinite number of spatio-temporal firing interpretations can be used to represent the same underlying computation. As a result, at two different instances a Boolean function can be computed differently by an active element machine. This substantially increases the AEM's resistance to reverse engineering and apprehension of the purpose of an AEM program. This enables a computer program instruction to be executed differently at different instances. In some embodiments, these different instances are at different times. In some embodiments, these different instances of computing the program instruction are executed by different collections of active elements and connections in the machine. Some embodiments use random active element machine firing interpretations to compute a Boolean function differently at two different instances.

E. Incomputability is used instead of complexity. Incomputability means that a general Turing machine algorithm can not unlock or solve an incomputable problem. This means that a digital computer program can not solve an incomputable problem. This creates a superior level of computational security.

F. Randomness in the AEM computing model. Because the AEM Interpretation approach relies on quantum randomness to dynamically generate random firing patterns, the AEM implementing this technique is no longer subject to current computability theory that assumes the Turing machine or register machine as the computing model. This means that prior art methods and their lack of solutions for malware that depend on Turing's halting problem and undecidability no longer apply to the AEM in this context. This is another aspect of the AEM's non-Turing behavior (i.e. beyond a digital computer's capabilities) that provides useful novel cybersecurity capabilities.

In some embodiments, the quantum randomness utilized with the AEM helps create a more powerful computational procedure in the following way. An active element machine (AEM) that uses quantum randomness can deterministically execute a universal Turing machine (i.e. digital computer program that can execute any possible digital computer program) such that the firing patterns of the AEM are Turing incomputable. An active element machine (AEM) that uses quantum randomness deterministically executes digital computer instructions such that the firing patterns of the active element machine are Turing incomputable. This means that this security method will work for any digital computer program and the capability works for any digital computer hardware/software implementation and for digital computer programs written in C, C++, JAVA, Fortran, Assembly Language, Ruby, Forth, Haskell, RISC machine instructions (digital computer machine instructions, JVM (Java virtual machine), Python and other digital computer languages.

Register machine instructions, Turing machine or digital computer instructions can be executed with active element machine instructions where it is Turing incomputable to understand what the active element machine computation is doing. In these embodiments, the active element machine computing behavior is non-Turing. This enhances the capability of a computational procedure: it secures the computational process (new secure computers) and helps protect a computation from malware.

Why is now a good time?

a. It was recently discovered that an Active Element machine can exhibit non-Turing dynamical behavior. The use of prime directed edge complexity was discovered. Every Turing machine (digital computer program) has a finite prime directed edge complexity. (See 4.20 and 4.23.) An active element machine that has unbounded prime directed edge complexity can be designed using physical randomness. For example, the physical or quantum randomness can be realized with quantum optics or quantum effects in a semiconductor or another quantum phenomena.

b. The Meta command was discovered which enables the AEM to change its program as execution proceeds. This enables the machine to compute the same computational operation in an infinite number of ways and makes it conducive to machine learning and self-repair. The AEM can compute with a language that randomly evolves while the AEM program is executing.

c. It was recently realized that Active Element machine programs can be designed that are connected (in terms of topology), making them resistant to tampering and hijacking.

d. When a Turing machine or register machine (digital computer) executes an unbounded (non-halting) computation, the long term behavior of the program has recurrent points. This demonstrates the machine's predictable computing behavior which creates weaknesses and attack points for malware to exploit. This recurrent behavior in Turing machine and register machine is described in the section titled IMMORTAL ORBIT and RECURRENT POINTS.

e. Randomness can be generated from physical processes using quantum phenomena i.e. quantum optics, quantum tunneling in a semiconductor or other quantum phenomena. Using quantum randomness as a part of the active element machine exhibits non-Turing computing behavior. This non-Turing computing behavior generates random AEM firing interpretations that are difficult for malware to comprehend.

What is novel about the new applications that can be built?

In some embodiments, an AEM can execute on current computer hardware and in some embodiments is augmented. These novel methods using an AEM are resistant to hackers and malware apprehending the purpose of AEM program's computations and in terms of sabotaging the AEM program's purpose; sabotaging a computation's purpose is analogous to a denial of service or distributed denial of service attack. The machine has computing performance that is orders of magnitude faster when implemented with hardware that is specifically designed for AEM computation. The AEM is useful in applications where reliability, security and performance are of high importance: protecting and reliably executing the Domain Name Servers, securing and running critical infrastructure such as the electrical grid, oil refineries, pipelines, irrigation systems, financial exchanges, financial institutions and the cybersecurity system that coordinates activities inside institutions such as the government.

BRIEF SUMMARY OF PRIOR ART COMPUTING MODELS

For completeness, a brief introduction to Turing machines is presented a later section. In [32], Alan Turing introduces the Turing Machine, which is a basis for the current digital computer. Sturgis and Shepherdson present the register machine in [31] and demonstrate the register machine's computational equivalence to the Turing machine: a Turing machine can compute a function in a finite number of steps if and only if a register machine can also compute this function in a finite number of steps. The works [7], [20], [21], [22] and [24] cover computability where other notions of computation equivalent to the Turing machine are also described.

In [23], McCulloch and Pitts present one of the early alternative computing models influenced by neurophysiology. In [27], Rosenblatt presents the perceptron model, which has a fixed number of perceptrons and has no feedback (cycles) in its computation. In [25], Minsky and Papert mathematically analyze the perceptron model and attempt to understand serial versus parallel computation by studying the capabilities of linear threshold predicates. In [16], Hopfield shows how to build a content addressable memory with neural networks that use feedback and where each neuron has two states. The number of neurons and connections are fixed during the computation. In [17], Hopfield presents an analog hardware neural network to perform analog computation on the Traveling-Salesman problem, which is NP-complete [12]. Good, suboptimal solutions to this problem are computed by the analog neural network within an elapsed time of only a few neural time constants.

In [18], Hopfield uses time to represent the values of variables. In the conclusion, he observes that the technique of using time delays is similar to that of using radial basis functions in computer science.

In [15], Hertz et al. discuss the Hopfield model and various computing models that extend his work. These models describe learning algorithms and use statistical mechanics to develop the stochastic Hopfield model. They use some statistical mechanics techniques to analyze the Hopfield model's memory capacity and the capacity of the simpler perceptron model.

For early developments on quantum computing models, see [2], [3], [9], [10], [21] and [22]. In [29], Shor discovers a quantum algorithm showing that prime factorization can be executed on quantum computers in polynomical time (i.e. considerably faster than any known classical algorithm). In [13], Grover discovers a quantum search algorithm among n objects that can be completed in cn^(0.5) computational steps.

In [8], Deutsch argues that there is a physical assertion in the underlying Church-Turing hypothesis: Every finitely realizable physical system can be perfectly simulated by a universal model computing machine operating by finite means. Furthermore, Deutsch presents a quantum generalization of the class of Turing machines: a universal quantum computer that covers quantum parallelism and shows an increase in computing speed. This universal quantum computer does not demonstrate the computation of non-Turing computable functions. For the most part, these prior results on computing models have studied the model's speed of computation, memory capacity, learning ability or have demonstrated that a particular computing model is equivalent to the Turing machine (digital computer)—in terms of computability (see [7] pages 10-12).

SUMMARY OF METHODS

a. AEM representations are randomly generated that are Turing incomputable to determine their computational purpose.

b. AEM representations are created that are also topologically connected.

c. AEM parallelism is used to solve computationally difficult tasks as shown in the section titled An AEM Program Computes a Ramsey Number.

d. Turing machine computation (digital computer computation) is topologically disconnected as shown by the affine map correspondence in 2.25.

Synthesis of Multiple Methods In some embodiments, multiple methods are used and the solution is a synthesis of some of the following methods, A-E. A. An AEM program—with input active elements fired according to b₁ b₂ . . . b_(m)—accepts [b₁ b₂ . . . b_(m)] if active elements E₁, E₂ . . . , E_(n) exhibit a set or sequence of firing patterns. In some embodiments, this sequence of firing patterns has Turing incomputable interpretations using randomness. B. AEM programs are created with an unbounded prime edge complexity. Turing and register machine programs have a finite prime directed edge complexity as shown in the section titled Prime Edge Complexity, Periodic Points & Repeating State Cycles. C. AEM programs are created with no recurrent points when computation is unbounded with respect to time. This is useful for cybersecurity as it helps eliminate weaknesses for malware to exploit. When a Turing machine or register machine (digital computer) executes an unbounded (non-halting) computation, the long term behavior of the program has recurrent points. The recurrent behavior in a digital computer is described in the section titled Immortal Orbit and Recurrent Points. D. Multiple AEM firing patterns are computed concurrently and then one can be selected according to an interpretation executed by a separate AEM machine. The AEM interpretation is kept hidden and changes over time. In some embodiments, evolutionary methods using randomness may help build AEMs that utilize incomputability and topological connectedness in their computations. E. In some embodiments, AEMs programs represent the Boolean operations in a digital computer using multiple spatio-temporal representations, which is further described in the detailed description. In some embodiments, level set methods on random AEM firing interpretations may be used that do not use Boolean functions. This enables a digital computer program instruction to be executed differently at different instances. In some embodiments, these different instances are at different times. In some embodiments, these different instances of computing the program instruction are executed by different collections of active elements and connections in the active element machine. F. In some embodiments, the parallel computing speed increase of an AEM is substantial. As described in the section titled An AEM Program Computes a Ramsey Number, an AEM program is shown that computes a Ramsey number using the parallelism of the AEM. The computation of Ramsey numbers is an NP-hard problem [12].

DETAILED DESCRIPTION

Although various embodiments of the invention may have been motivated by various deficiencies with the prior art, which may be discussed or alluded to in one or more places in the specification, the embodiments of the invention do not necessarily address any of these deficiencies. In other words, different embodiments of the invention may address different deficiencies that may be discussed in the specification. Some embodiments may only partially address some deficiencies or just one deficiency that may be discussed in the specification, and some embodiments may not address any of these deficiencies.

Active Element Machine Description

An active element machine is composed of computational primitives called active elements. There are three kinds of active elements: Input, Computational and Output active elements. Input active elements receive information from the environment or another active element machine. This information received from the environment may be produced by a physical process.

In some embodiments, information from the environment may come from the physical process of photons originating from sunlight or other kinds of light traveling through space. In some embodiments, information from the environment may come from the physical process of sound. The sounds waves may be received by a sensor or transducer that causes one or more input elements to fire. In some embodiments, the acoustic transducer may be a part of the input elements and each input element may be more sensitive to a range of sound frequencies. In some embodiments, the sensor(s) or transducer(s) may be a part of one or more of the input elements and each input element may be more sensitive to a range of light frequencies analogous to the cones in the retina.

In some embodiments, information from the environment may come from the physical process of molecules present in the air or water. In some embodiments, sensor(s) or transducer(s) may be sensitive to particular molecules diffusing in the air or water, which is analogous to the molecular receptors in a person's nose. For example, one or more input elements may fire if a particular concentration of cinnamon molecules are detected by olfactory sensor(s).

In some embodiments, the information from the environment may originate from the physical process of pressure. In some embodiments, pressure information is transmitted to one or more of the input elements. In some embodiments, the sensor(s) that are a part of the input elements or connected to the input elements may be sensitive to pressure, which is analogous to a person's skin. In some embodiments, sensor sensitive to heat may be a part of the input elements or may be connected to the input elements. This is analogous to a person's skin detecting temperature.

Computational active elements receive messages from the input active elements and other computational active elements firing activity and transmit new messages to computational and output active elements. The output active elements receive messages from the input and computational active elements firing activity. Every active element is active in the sense that each one can receive and transmit messages simultaneously.

Each active element receives messages, formally called pulses, from other active elements and itself and transmits messages to other active elements and itself. If the messages received by active element E_(i) at the same time sum to a value greater than the threshold and E_(i)'s refractory period has expired, then active element E_(i) fires. When an active element E_(i) fires, it sends messages to other active elements.

Let Z denote the integers. Define the extended integers as

={m+kdT: m, kεZ and dT is a fixed infinitesimal}. For more on infinitesimals, see [26] and [14]. The extended integers can also be expressed using the correspondence m+ndT

(m, n) where (m, n) lies in Z×Z. Then use the dictionary order (m, n)<(k, l) if and only if (m<k) OR (m=k AND n<l). Similarly, m+ndT<k+ldT if and only if (m<k) OR (m=k AND n<1).

Machine Architecture

Γ, Ω, and Δ are index sets that index the input, computational, and output active elements, respectively. Depending on the machine architecture, the intersections Γ∩Ω and Ω∩Δ can be empty or non-empty. A machine architecture, denoted as

(I,

, O), consists of a collection of input active elements, denoted as I={E_(i): iεΓ}; a collection of computational active elements

={E_(i): iεΩ}; and a collection of output active elements O={E_(i): iεΔ}.

Each computational and output active element, E_(i), has the following components and properties.

A threshold θ_(i)

A refractory period r_(i) where r_(i)>0.

A collection of pulse amplitudes {A_(ki): kεΓ∪Ω}.

A collection of transmission times {τ_(ki): kεΓ∪Ω}, where τ_(ki)>0 for all kεΓ∪Ω.

A function of time, Ψ_(i)(t), representing the time active element E_(i) last fired. Ψ_(i)(t)=sup{s: s<t and g_(i)(s)=1}, where g_(i)(s) is the output function of active element E_(i) and is defined below. The sup is the least upper bound.

A binary output function, g_(i)(t), representing whether active element E_(i) fires at time t. The value of g_(i)(t)=1 if ΣA_(ki)(t)>0; where the sum ranges over all kεΓ∪Ω and t≧Ψ_(i)(t)+r_(i). In all other cases, g_(i)(t)=0. For example, g_(i)(t)=0, if t<Ψ_(i)(t)+r_(i).

A set of firing times of active element E_(k) within active element E_(i)'s integrating window, W_(ki)(t)={s: active element E_(k) fired at time s and 0≦t−s−τ_(ki)<ω_(ki)}. Let |W_(ki)(t)| denote the number of elements in the set W_(ki)(t). If W_(ki)(t)=Ø, then |W_(ki)(t)|I=0.

A collection of input functions, {(φ_(ki): kεΓ∪Ω}, each a function of time, and each representing pulses coming from computational active elements, and input active elements. The value of the input function is computed as φ_(ki)(t)=|W_(ki)(t)|A_(ki)(t).

The refractory periods, transmission times and pulse widths are positive integers; and pulse amplitudes and thresholds are integers. These parameters are a function of i.e. θ_(i)(t), r_(i)(t), A_(ki)(t), ω_(ki)(t), τ_(ki)(t). The time t is an element of the extended integers

.

Input active elements that are not computational active elements have the same characteristics as computational active elements, except they have no inputs φki coming from active elements in this machine. In other words, they don't receive pulses from active elements in this machine. Input active elements are assumed to be externally firable. An external source such as the environment or an output active element from another distinct machine

(I,

, O) can cause an input active element to fire. The input active element can fire at any time as long as the current time minus the time the input active element last fired is greater than or equal to the input active element's refractory period.

An active element E_(i) can be an input active element and a computational active element. Similarly, an active element can be an output active element and a computational active element. Alternatively, when an output active element E_(i) is not a computational active element, where iεΔ−Ω, then E_(i) does not send pulses to active elements in this machine.

Some notions of the machine architecture are summarized. If g_(i)(s)=1, this means active element E_(i) fired at time s. The refractory period r_(i) is the amount of time that must elapse after active element E_(i) just fired before E_(i) can fire again. The transmission time τ_(ki) is the amount of time it takes for active element E_(i) to find out that active element E_(k) has fired. The pulse amplitude A_(ki) represents the strength of the pulse that active element E_(k) transmits to active element E_(i) after active element E_(k) has fired. After this pulse reaches E_(i), the pulse width ω_(ki) represents how long the pulse lasts as input to active element E_(i). At time s, the connection from E_(k) to E_(i) represents the triplet (A_(ki)(s), ω_(ki)(S), τ_(ki)(S)). If A_(ki)=0, then there is no connection from active element E_(k) to active element E.

Refractory Period

In an embodiment, each computational element and output element has a refractory period r_(i), where r_(i)>0, which is a period of time that must elapse after last sending a message before it may send another message. In other words, the refractory period, r_(i), is the amount of time that must elapse after active element E_(i) just fired and before active element E₁ can fire again. In an alternative embodiment, refractory period r_(i) could be zero, and the active element could send a message simultaneously with receiving a message and/or could handle multiple messages simultaneously.

Message Amplitude and Width

In an embodiment, each computational element and output element may be associated with a collection of message amplitudes, {A_(ki)}_(k Γ∪Λ), where the first of the two indices k and i denote the active element from which the message associated with amplitude A_(ki) is sent, and the second index denotes the active element receiving the message. The amplitude, A_(ki), represents the strength of the message that active element E_(k) transmits to active element E_(i) after active element E_(k) has fired. There are many different measures of amplitude that may be used for the amplitude of a message. For example, the amplitude of a message may be represented by the maximum value of the message or the root mean square height of the message. The same message may be sent to multiple active elements that are either computational elements or output elements, as indicated by the subscript k Γ∪Λ. However, each message may have a different amplitude A_(ki). Similarly, each message may be associated with its own message width, {ω_(ki)}_(k Γ∪Λ), sent from active element E_(i) to E_(k), where ω_(ki)>0 for all Γ∪Λ. After a message reaches active E_(i), the message width ω_(ki) represents how long the message lasts as input to active element E_(i).

Threshold

In an embodiment, any given active element may be capable of sending and receiving a message, in response to receiving one or more messages, which when summed together, have an amplitude that is greater than a threshold associated with the active element. For example, if the messages are pulses, each computational and output active element, E_(i), may have a threshold, θ_(i), such that when a sum of the incoming pulses is greater than the threshold the active element fires (e.g., sends an output message). In an embodiment, when a sum of the incoming messages is lower than the threshold the active element does not fire. In another embodiment, it is possible to set the active element such that the active element fires when the sum of incoming messages is lower than the threshold; and when the sum of incoming messages is higher than the threshold, the active element does not fire.

In still another embodiment, there are two numbers α and θ where α≦θ and such that if the sum of the incoming messages lie in [α, θ], then the active element fires, but the active element does not fire if the sum lies outside of [α, θ]. In a variation of this embodiment, the active element fires if the sum of the incoming messages does not lie in [α, θ] and does not fire if the sum lies in [α, θ].

In another embodiment, the incoming pulses may be combined in other ways besides a sum. For example, if the product of the incoming pulses is greater than the threshold the active element may fire. Another alternative is for the active element to fire if the maximum of the incoming pulses is greater than the threshold. In still another alternative, the active element fires if the minimum of the incoming pulses is less than the threshold. In even another alternative if the convolution of the incoming pulses over some finite window of time is greater than the threshold, then the active element may fire.

Transmission Time

In an embodiment, each computational and output element may be associated with collection of transmission times, {τ_(ki)}_(k Γ∪Λ), where τ_(ki)>0 for all k Γ∪Λ, which are the times that it takes a message to be sent from active element E_(k) to active element E_(i). The transmission time, τ_(ki), is the amount of time it takes for active element E_(i) to find out that active element E_(k) has fired. The transmission times, τ_(ki), may be chosen in the process of establishing the architecture.

Firing Function

In an embodiment, each active element is associated with a function of time, ψ_(i)(t), representing the time t at which active element E_(i) last fired. Mathematically, the function of time can be defined as ψ_(i)(t)=supremum {s R: s<t AND g_(i)(s)=1}. The function ψ_(i)(t) always has the value of the last time that the active element fired. In general, throughout this specification the variable t is used to represent the current time, while in contrast s is used as variable of time that is not necessarily the current time.

Set of Firing Times and the Integrating Window

In an embodiment, each active element is associated with a function of time Ξ_(ki)(t), which is a set of recent firing times of active element E_(k) that are within active element E_(i)'s integrating window. In other words, the set of firing times Ξ_(ki)(t)={s R: active element k fired at time s and 0≦t−s−τ_(ki)<ω_(ki)}. The integrating window is a duration of time during which the active element accepts messages. The integrating window may also be referred to as the window of computation. Other lengths of time could be chosen for the integrating window. In contrast to ψ_(i) (t), Ξ_(k); (t) is not a function, but a set of values. Also, where as ψ_(i)(t) has a value as long as active element E_(i) fired at least once, Ξ_(ki)(t) does not have any values (is an empty set) if the last time that active element E_(i) fired is outside of the integrating window. In other words, if there are no firing times, s, that satisfy the inequality 0≦t−s−τ_(ki)<ω_(ki), then Ξ_(ki)(t) is the empty set. Let |Ξ_(ki)(t)| denote the number of elements in the set Ξ_(ki)(t). If Ξ_(ki)(t) is the empty set, then |Ξ_(ki)(t)|=0. Similarly, if Ξ_(ki)(t) has only one element in it then |Ξ_(ki)(t)|=1.

Input Function

In an embodiment, each input element and output element may have associated with it a collection of input functions, {Ø_(ki)(t)}_(k Γ∪Λ). Each input function may be a function of time, and may represent messages coming from computational elements and input elements. The value of input function Ø_(ki)(t) is given by Ø_(ki)(t)=|Ξ_(ki)(t)|A_(ki), because each time a message from active element E_(k) reaches active element E_(i), the amplitude of the message is added to the last message. The number of messages inside the integrating window is the same as the value of |Ξ_(ki)(t)|. Since for a static machine the amplitude of the message sent from active element k to i is always the same value, A_(ki), therefore, the value Ø_(ki)(t) equals |Ξ_(ki)(t)|A_(ki).

Input elements that are not computational elements have the same characteristics as computational elements, except they have no input functions, Ø_(ki)(t), coming from active elements in this machine. In other words, input elements do not receive messages from active elements in the machine with which the input element is associated. In an embodiment, input elements are assumed to be externally firable. An externally firable element is an element that an external element or machine can cause to fire. In an embodiment, an external source such as the environment or an output element from another distinct machine, M′(J′; E′;D′) can cause an input element to fire. An input element can fire at any time as long as this time minus the time the input element last fired is greater than or equal to the input element's refractory period.

Output Function

An output function, g_(i)(t), may represent whether the active element fires at time t. The function g_(i)(t) is given by

${g_{i}(t)} = \left\{ \begin{matrix} 1 & {{{if}\mspace{14mu} {\sum\limits_{k \in {\Gamma \; U\; \Lambda}}Ø_{ki}}} > {{\theta_{i}\mspace{14mu} {AND}\mspace{11mu} t} - {\psi_{i}(t)}} \geq r_{i}} \\ 0 & {{otherwise}.} \end{matrix} \right.$

In other words, if the sum of the input functions Ø_(ki) (t) is greater than the threshold, θ_(i), and time t is greater than or equal to the refractory period, r_(i), plus the time, ψ_(i)(t), that the active element last fired, then the active element E_(i) fires, and g_(i)(t)=1. If g_(i)(t₀)=1, then active element E_(i) fired at time t₀.

The fact that in an embodiment, output elements do not send messages to active elements in this machine is captured formally by the fact that the index k for the transmission times, message widths, message amplitudes, and input functions lies in Γ∪Λ and not in Δ in that embodiment.

Connections

The expression “connection” from k to i represents the triplet (A_(ki); ω_(ki); τ_(ki)). If A_(ki)=0, then there is no connection from active element E_(k) to active element E_(i). If A_(ki)≠0, then there is a non-zero connection from active element E_(k) to active element E_(i). In any given embodiment the active elements may have all of the above properties, only one of the above properties, or any combination of the above properties. In an embodiment, different active elements may have different combinations of the above properties. Alternatively, all of the active elements may have the same combination of the above properties.

Active Element Machine Programming Language

This section shows how to program an active element machine and how to change the machine architecture as program execution proceeds. It is helpful to define a programming language, influenced by S-expressions. There are five types of commands: Element, Connection, Fire, Program and Meta.

Syntax 1. AEM Program

In Backus-Naur form, an AEM program is defined as follows.

<AEM_program> ::= <cmd_sequence> <cmd_sequence> ::= “” | <AEM_cmd><cmd_sequence> | <program_def><cmd_sequence> <AEM_cmd> ::= <element_cmd> | <fire_cmd> | <meta_cmd> | <cnct_cmd> | <program_cmd> <ename> ::= “” | <int> | <symbol> <symbol> ::= <symbol_string> | (<ename> . . . <ename>) <symbol_string> ::= “” | <char_symbol><str_tail> <str_tail> ::= “” | <char_symbol><str_tail> | 0<str_tail> | <pos_int><str_tail> <char_symbol> ::= <letter> | <special_char> <letter> ::= <lower_case> | <upper_case> <lower_case> ::= a | b | c | d | e | f | g | h | i | j | k | l | m | n | o | p | q | r | s | t | u | v | w | x | y | z <upper_case> ::= A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z <special_char> ::= “” | _(—)

These rules represent the extended integers, addition and subtraction.

<int> ::= <pos_int> | <neg_int> | 0 <neg_int> ::= − <pos_int> <pos_int> ::= <non_zero><digits> <digits> ::= <numeral> | <numeral><digits> <non_zero> ::= 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 <numeral> ::= “” | <non_zero> | 0 <aint> ::= <aint> <math_op> <d> | <d> <math_op> <aint> | <d> <math_op> ::= + | − <d> ::= <int> | <symbol_string> | <infinitesimal> <infinitesimal> ::= dT

Element Command.

An Element command specifies the time when an active element's values are updated or created. This command has the following Backus-Naur syntax.

<element_cmd> ::= (Element (Time <aint>) (Name <ename>) (Threshold <int>) (Refractory <pos_int>) (Last <int>))

The keyword Time indicates the time value s at which the element is created or updated. In some embodiments the time value s is an extended integer. If the name symbol value is E, the keyword Name tags the name E of the active element. The keyword Threshold tags the threshold θ_(E)(s) assigned to E. Refractory indicates the refractory value r_(E)(s). The keyword Last tags the last time fired value Ψ(s). Sometimes the time value, name value, threshold value, refractory value, or last time fired value are referred to as parameter values.

Below is an example of an element command.

(Element (Time 2) (Name H) (Threshold −3) (Refractory 2) (Last 0))

At time 2, if active element H does not exist, then it is created. Active element H has its threshold set to −3, its refractory period set to 2, and its last time fired set to 0. After time 2, active element H exists indefinitely with threshold=−3, refractory=2 until a new Element command whose name value H is executed at a later time; in this case, the Threshold, Refractory and Last values specified in the new command are updated.

Connection Command.

A Connection command creates or updates a connection from one active element to another active element. This command has the following Backus-Naur syntax.

<cnct_cmd> ::= (Connection (Time <aint>) (From <ename>) (To <ename>) [(Amp <int>) (Width <pos_int>) (Delay <pos_int>)])

The keyword Time indicates the time value s at which the connection is created or updated. In some embodiments the time value s is an extended integer. The keyword From indicates the name F of the active element that sends a pulse with these updated values. The keyword To tags the name T of the active element that receives a pulse with these updated values. The keyword Amp indicates the pulse amplitude value A_(FT)(s) that is assigned to this connection. The keyword Width indicates the pulse width value ω_(FT)(s). In some embodiments the pulse width value ω_(FT)(s) is an extended integer. The keyword Delay tags the transmission time τ_(FT)(s). In some embodiments the transmission time τ_(FT)(s) is an extended integer. Sometimes the time value, from name, to name, pulse amplitude value, pulse width value, or transmission time value are referred to as parameter values.

When the AEM clock reaches time s, F and T are name values that must be the name of an element that already has been created or updated before or at time s. Not all of the connection parameters need to be specified in a connection command. If the connection does not exist beforehand and the Width and Delay values are not specified appropriately, then the amplitude is set to zero and this zero connection has no effect on the AEM computation. Observe that the connection exists indefinitely with the same parameter values until a new connection is executed at a later time between From element F and To element T.

The following is an example of a connection command.

(Connection (Time 2) (From C) (To L) (Amp −7) (Width 1) (Delay 3))

At time 2, the connection from active element C to active element L has its amplitude set to −7, its pulse width set to 1, and its transmission time set to 3.

Fire Command.

The Fire command has the following Backus-Naur syntax.

<fire_cmd>::=(Fire (Time <aint>) (Name <ename>))

The Fire command fires the active element indicated by the Name tag at the time indicated by the Time tag. Sometimes the time value and name value are referred to as parameter values of the fire command. In some embodiments, the fire command is used to fire input active elements in order to communicate program input to the active element machine. An example is (Fire (Time 3) (Name C)), which fires active element C at t=3.

Program Command.

The Program command is convenient when a sequence of commands are used repeatedly. This command combines a sequence of commands into a single command. It has the following definition syntax.

<program_def> ::= (Program <pname> [(Cmds <cmds>)] [(Args <args>)] <cmd_sequence> ) <pname> ::= <ename> <cmds> ::= <cmd_name> | <cmd_name><cmds> <cmd_name> ::= Element | Connection | Fire | Meta | <pname> <args> ::= <symbol> | <symbol><args>

The Program command has the following execution syntax.

<program_cmd> ::= (<pname> [(Cmds <cmds>)] [(Args <args_cmd>)] ) <args_cmd> ::= <ename> | <ename><args_cmd>

The FireN program is an example of definition syntax.

(Program FireN (Args t E) (Element (Time 0) (Name E)(Refractory l)(Threshold 1) (Last 0)) (Connection (Time 0) (From E) (To E)(Amp 2)(Width 1) (Delay 1)) (Fire (Time 1) (Name E)) (Connection (Time t+1) (From E) (To E) (Amp 0)) )

The execution of the command (FireN (Args 8 E1)) causes element E1 to fire 8 times at times 1, 2, 3, 4, 5, 6, 7, and 8 and then E1 stops firing at time=9.

Keywords Clock and dT

The keyword clock evaluates to an integer, which is the current active element machine time. clock is an instance of <ename>. If the current AEM time is 5, then the command

(Element (Time clock) (Name clock) (Threshold 1) (Refractory 1) (Last −1)) is executed as

(Element (Time 5) (Name 5) (Threshold 1) (Refractory 1) (Last −1))

Once command (Element (Time clock) (Name clock) (Threshold 1) (Refractory 1) (Last −1)) is created, then at each time step this command is executed with the current time of the AEM. If this command is in the original AEM program before the clock starts at 0, then the following sequence of elements named 0, 1, 2, . . . will be created.

(Element (Time 0) (Name 0) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 1) (Name 1) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 2) (Name 2) (Threshold 1) (Refractory 1) (Last −1)) . . .

The keyword dT represents a positive infinitesimal amount of time. If m and n are integers and 0≦m<n, then mdT<ndT. Furthermore, dT>0 and dT is less than every positive rational number. Similarly, −dT<0 and −dT is greater than every negative rational number. The purpose of dT is to prevent an inconsistency in the description of the machine architecture. For example, the use of dT helps remove the inconsistency of a To element about to receive a pulse from a From element at the same time that the connection is removed.

Meta Command.

The Meta command causes a command to execute when an element fires within a window of time. This command has the following execution syntax.

<meta_cmd> ::= (Meta (Name <ename>)[<win_time>] <AEM_cmd>) <win_time> ::= (Window <aint> <aint>)

To understand the behavior of the Meta command, consider the execution of

(Meta (Name E) (Window l w) (C (Args t a))

where E is the name of the active element. The keyword Window tags an interval i.e. a window of time. l is an integer, which locates one of the boundary points of the window of time. Usually, w is a positive integer, so the window of time is [l, l+w]. If w is a negative integer, then the window of time is [l+w, l].

The command C executes each time that E fires during the window of time, which is either [l, l+w] or [l+w, l], depending on the sign of w. If the window of time is omitted, then command C executes at any time that element E fires. In other words, effectively l=−∞ and w=∞. Consider the example where the FireN command was defined before.

(FireN (Args 8 E1)) (Meta (Name E1) (Window 1 5) (C (Args clock a b)) )

Command C is executed 6 times with arguments dock, a, b. The firing of E1 triggers the execution of command C.

In regard to the Meta command, the following assumption is analogous to the Turing machine tape being unbounded as Turing program execution proceeds. During execution of a finite active element program, an active element can fire and due to one or more Meta commands, new elements and connections can be added to the machine. As a consequence, at any time the active element machine only has a finite number of computing elements and connections but the number of elements and connections can be unbounded as a function of time as the active element program executes.

Active Element Machine Computation

In a prior section, the firing patterns of active elements are used to represent the computation of a boolean function. In the next three definitions, firing representations, machine computation and interpretation are defined.

Firing Representation

Consider active element E_(i)'s firing times in the interval of time W=[t₁, t₂]. Let s₁ be the earliest firing time of E_(i) lying in W, and s_(n) the latest firing time lying in W. Then E_(i)'s firing sequence

(E_(i), W)=[s₁, . . . , s_(n)]={sεW: g_(i)(s)=1} is called a firing sequence of the active element E, over the window of time W. From active elements {E₁, E₂, . . . , E_(n)} create the tuple (

(E₁, W),

(E₂, W), . . . ,

(E_(n), W)) which is called a firing representation of the active elements {E₁, E₂, . . . , E_(n)} within the window of time W.

At the machine level of interpretation, firing representations (firing patterns) express the input to, the computation of, and the output of an active element machine. At a more abstract level, firing representations can represent an input symbol, an output symbol, a sequence of symbols, a spatio-temporal pattern, a number, or even a family of program instructions for another computing machine.

Sequence of Firing Representations.

Let W₁, . . . , W_(n) be a sequence of time intervals. Let

(

, W₁)=(

(E₁, W₁),

(E₂, W₁), . . . ,

(E_(n), W₁)) be a firing representation of active elements

={E₁, . . . , E_(n)} over the interval W₁. In general, let

(E, W_(k))=(

(E₁, W_(k)),

(E₂, W_(k)), . . .

(E_(n), W_(k))) be a firing representation over the interval of time W_(k). From these, a sequence of firing representations, [

(

, W₁),

(

, W₂), . . . ,

(

, W_(n))] is created.

Machine Computation

Let [

(

, W₁),

(

, W₂), . . .

(

, W_(n))] be a sequence of firing representations. [

(

, S₁),

(

, S₂), . . . ,

(

, S_(m))] is some other sequence of firing representations. Suppose machine architecture

(I, E, O) has input active elements/fire with the pattern [

(

, S₁),

(

, S₂), . . . ,

(

, S_(m))] and consequently

's output active elements O fire according to [

(

, W₁),

(

, W₂), . . . ,

(

, W_(n))]. In this case, the machine M computes [

(

, W₁),

(

, W₂), . . . ,

(

, W_(n))] from [

(

, S₁),

(

, S₂), . . . ,

(

, S_(m))].

An active element machine is an interpretation between two sequences of firing representations if the machine computes the output sequence of firing representations from the input sequence of firing representations.

Concurrent Generation of AEM Commands

This section shows embodiments pertaining to two or more commands about to set parameter values of the same connection or same element at the same time. Consider two or more connection commands, connecting the same active elements, that are generated and scheduled to execute at the same time.

(Connection (Time t) (From A) (To B) (Amp 2) (Width 1) (Delay 1)) (Connection (Time t) (From A) (To B) (Amp −4) (Width 3) (Delay 7))

Then the simultaneous execution of these two commands can be handled by defining the outcome to be equivalent to the execution of only one connection command where the respective amplitudes, widths and transmission times are averaged.

(Connection (Time t) (From A) (To B) (Amp −1) (Width 2) (Delay 4))

In the general case, for n connection commands

(Connection (Time t) (From A) (To B) (Amp a1) (Width w1) (Delay s1)) (Connection (Time t) (From A) (To B) (Amp a2) (Width w2) (Delay s2)) . . . (Connection (Time t) (From A) (To B) (Amp an)(Width wn)(Delay sn)) resolve these to the execution of one connection command

(Connection (Time t) (From A) (To B) (Amp a) (Width w) (Delay s))

where a, w and s are defined based on the application. These embodiments can be implemented in active element machine software, AEM hardware or a combination of AEM hardware and software.

For some embodiments of the AEM, averaging the respective amplitudes, widths and transmission times is useful.

a=(a1+a2+ . . . +an)/n

w=(w1+w2+ . . . +wn)/n

s=(s1+s2+ . . . +sn)/n

For embodiments that use averaging, they can be implemented in active element machine software, AEM hardware or a combination of AEM hardware and software.

For some embodiments, when there is noisy environmental data fed to the input elements and amplitudes, widths and transmission times are evolved and mutated, extremely large (in absolute value) amplitudes, widths and transmission times can arise that skew an average function. In these embodiments, computing the median of the amplitudes, widths and delays provides a simple method to address skewed amplitude, width and transmission time values.

a=median(a1,a2, . . . ,an)

w=median(w1,w2, . . . ,wn)

s=median(s1,s2, . . . ,sn)

Another alternative embodiment adds the parameter values.

a=a1+a2+ . . . +an

w=w1+w2+ . . . +wn

s=s1+s2+ . . . +sn

Similarly, consider when two or more element commands—that all specify the same active element E—are generated and scheduled to execute at the same time.

(Element (Time t) (Name E) (Threshold h1) (Refractory r1) (Last s1)) (Element (Time t) (Name E) (Threshold h2) (Refractory r2) (Last s2)) . . . (Element (Time t) (Name E) (Threshold hn) (Refractory rn) (Last sn)) resolve these to the execution of one element command,

(Element (Time t) (Name E) (Threshold h) (Refractory r) (Last s))

where h, r and s are defined based on the application. Similar to the connection command, for theoretical studies of the AEM, the threshold, refractory and last time fired values can be averaged.

h=(h1+h2+ . . . +hn)/n

r=(r1+r2+ . . . +rn)/n

s=(s1+s2+ . . . +sn)/n

In autonomous embodiments, where evolution of parameter values occurs, the median can also help address skewed values in the element commands.

h=median(h1,h2, . . . hn)

r=median(r1,r2, . . . rn)

s=median(s1,s2, . . . sn)

Another alternative is to add the parameter values.

h=h1+h2+ . . . +hn

r=r1+r2+ . . . +rn

s=s1+s2+ . . . +sn

Rules A, B, and C resolve concurrencies pertaining to the Fire, Meta and Program commands. Rule A. If two or more Fire commands attempt to fire element E at time t, then element E is fired just once at time t.

Rule B. Only one Meta command can be triggered by the firing of an active element. If a new Meta command is created and it happens to be triggered by the same element E as a prior Meta command, then the old Meta command is removed and the new Meta command is triggered by element E.

Rule C. If a Program command is called by a Meta command, then the Program's internal Element, Connection, Fire and Meta commands follow the previous concurrency rules defined. If a Program command exists within a Program command, then these rules are followed recursively on the nested Program command.

An AEM Program Computes a Ramsey Number

This section shows how to compute a Ramsey number with an AEM program. Ramsey theory can be intuitively described as structure which is preserved under finite decomposition. Applications of Ramsey theory include computer science, including lower bounds for parallel sorting, game theory and information theory. Progress on determining the basic Ramsey numbers r(k, l) has been slow. For positive integers k and l, r(k, l) denotes the least integer n such that if the edges of the complete graph K_(n) are 2-colored with colors red and blue, then there always exists a complete subgraph K_(k) containing all red edges or there exists a subgraph K_(l) containing all blue edges. To put our slow progress into perspective, arguably the best combinatorist of the 20th century, Paul Erdos asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of r(5, 5) or they will destroy our planet. In this case, Erdos claims that we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose instead that they ask for r(6, 6). For r(6, 6), Erdos believes that we should attempt to destroy the aliens.

Theorem R. The Standard Finite Ramsey Theorem.

For any positive integers m, k, n, there is a least integer N(m, k, n) with the following property: no matter how we color each of the n-element subsets of S={1, 2, . . . , N} with one of k colors, there exists a subset Y of S with at least m elements, such that all n-element subsets of Y have the same color.

When G and H are simple graphs, there is a special case of theorem R. Define the Ramsey number r(G, H) to be the smallest N such that if the complete graph K_(N) is colored red and blue, either the red subgraph contains G or the blue subgraph contains H. (A simple graph is an unweighted, undirected graph containing no graph loops or multiple edges. In a simple graph, the edges of the graph form a set and each edge is a pair of distinct vertices.) In [10], S. A. Burr proves that determining r(G, H) is an NP-hard problem.

An AEM program is shown that solves a special case of Theorem 3. Similar embodiments can compute larger Ramsey numbers. Consider the Ramsey number where each edge of the complete graph K₆ is colored red or blue. Then there is always at least one triangle, which contains only blue edges or only red edges. In terms of the standard Ramsey theorem, this is the special case N(3, 2, 2) where n=2 since we color edges (i.e. 2-element subsets); k=2 since we use two colors; and m=3 since the goal is to find a red or blue triangle. To demonstrate how an AEM program can be designed to compute N(3, 2, 2)=6, an AEM program is shown that verifies N(3, 2, 2)>5.

The symbols B and R represent blue and red, respectively. Indices are placed on B and R to denote active elements that correspond to the K₅ graph geometry. The indices come from graph geometry. Let E={{1,2},{1,3},{1,4},{1,5}, {2,3},{2,4},{2,5}, {3,4},{3,5},{4,5}} denote the edge set of K₅.

The triangle set T={{1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}}. Each edge is colored red or blue. Thus, the red edges are {{1,2}, {1,5}, {2,3}, {3,4}, {4,5}} and the blue edges are {{1,3},{1,4},{2,4},{2,5},{3,5}}. Number each group of AEM commands for K₅, based on the group's purpose. This is useful because these groups will be used when describing the computation for K₆.

1. The elements representing red and blue edges are established as follows.

(Element (Time 0) (Name R_12) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name R_15) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name R_23) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name R_34) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name R_45) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_13) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_14) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_24) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_25) (Threshold 1) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_35) (Threshold 1) (Refractory 1) (Last −1))

2. Fire element R_(—)24 if edge {j, k} is red.

(Fire (Time 0) (Name R_12)) (Fire (Time 0) (Name R_15)) (Fire (Time 0) (Name R_23)) (Fire (Time 0) (Name R_34)) (Fire (Time 0) (Name R_45))

Fire element B_jk if edge {j, k} is blue where j<k.

(Fire (Time 0) (Name B_13)) (Fire (Time 0) (Name B_14)) (Fire (Time 0) (Name B_24)) (Fire (Time 0) (Name B_25)) (Fire (Time 0) (Name B_35))

3. The following Meta commands cause these elements to keep firing after they have fired once.

(Meta (Name R_jk) (Window 0 1) (Connection (Time 0) (From R_jk) (To R_jk) (Amp 2) (Width 1) (Delay 1))) (Meta (Name B_jk) (Window 0 1) (Connection (Time 0) (From B_jk) (To B_jk) (Amp 2) (Width 1) (Delay 1)))

4. To determine if a blue triangle exists on vertices {i, j, k}, where {i, j, k} ranges over T, three connections are created for each potential blue triangle.

(Connection (Time 0) (From B_ij) (To B_ijk) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 0) (From B_jk) (To B_ijk) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 0) (From B_ik) (To B_ijk) (Amp 2) (Width 1) (Delay 1))

5. To determine if a red triangle exists on vertex set {i, j, k}, where {i, j, k} ranges over T, three connections are created for each potential red triangle.

(Connection (Time 0) (From R_ij) (To R_ijk) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 0) (From R_jk) (To R_ijk) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 0) (From R_ik) (To R_ijk) (Amp 2) (Width 1) (Delay 1))

6. For each vertex set {i, j, k} in T, the following elements are created.

(Element (Time 0) (Name R_ijk) (Threshold 5) (Refractory 1) (Last −1)) (Element (Time 0) (Name B_ijk) (Threshold 5) (Refractory 1) (Last −1))

Because the threshold is 5, the element R_ijk only fires when all three elements R_ij, R_jk, R_ik fired one unit of time ago. Likewise, the element B_jjk only fires when all three elements B_jj, B_jk, B_jk fired one unit of time ago. From this, we observe that as of clock=3 i.e. 4 time steps, this AEM program determines that N(3, 2, 2)>5. This AEM computation uses

${{E} + {2{T}}} = {{\frac{5!}{{2!}{3!}} + {2\frac{5!}{{3!}{2!}}}} = {30\mspace{14mu} {active}\mspace{14mu} {{elements}.}}}$

Further, this AEM program creates and uses 3|T|+3|T|+|E|=70 connections.

For K₆, the edge set E={{1, 2}, {1, 3}, {1, 4}, {1, 5}, ≡1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}}. The triangle set T={{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1, 3, 4}, {1, 3, 5}, {1, 3, 6}, {1, 4, 5}, {1, 4, 6}, {1, 5, 6}, {2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 4, 5}, {2, 4, 6}, {2, 5, 6}, {3, 4, 5}, {3, 4, 6}, {3, 5, 6}, {4, 5, 6}1. For each 2-coloring of E, each edge is colored red or blue. There are 2^(|E|)2-colorings of E. For this graph,

${E} = \frac{6!}{{2!}{4!}}$

To build a similar AEM program, the commands in groups 1 and 2 range over every possible 2-coloring of E. The remaining groups 3, 4, 5 and 6 are the same based on the AEM commands created in groups 1 and 2 for each particular 2-coloring.

This AEM program verifies that every 2-coloring of E contains at least one red triangle or one blue triangle i.e. N(3, 2, 2)=6. There are no optimizations using graph isomorphisms made here. If an AEM language construct is used for generating all active elements for each 2-coloring of E at time zero, then the resulting AEM program can determine the answer in 5 time steps. One more time step is needed, 215 additional connections and one additional element to verify that every one of the 215 AEM programs is indicating that it found a red or blue triangle. This AEM program—that determines the answer in 5 time steps—uses 2^(|E|)(|E|+2|T|)+1 active elements and 2^(|E|)(3|T|+3|T|+|E|+1) connections, where |E|=15 and |T|=20. Some graph problems are related to the computation of Ramsey numbers.

A couple common graph problems are the traveling salesman problem and the traveling purchaser problem. (See <http://en.wikipedia.org/wiki/Traveling_salesman_problem> and <http://en.wikipedia.org/wiki/Traveling_purchaser_problem>.)

The traveling salesman problem can be expressed as a list of cities and their pairwise distances. The solution of the problem consists of finding the shortest possible tour that visits each city exactly once. Methods of solving the traveling salesman problem are useful for FedEx and other shipping companies where fuel costs and other shipping costs are substantial.

The traveling salesman problem has applications in planning, logistics, and the manufacture of microchips. Slightly modified, the traveling salesman problem appears as a sub-problem in many areas, such as DNA sequencing. In these applications, the concept city represents, for example, customers, soldering points, or DNA fragments, and the concept distance represents travelling times or cost, or a similarity measure between DNA fragments.

In embodiments similar to the computation of Ramsey numbers, the cities (customers, soldering points or DAN fragments) correspond to active elements and there is one connection between two cities for a possible tour that is being explored. In some embodiments, the thresholds of the two elements are used to account for the distance between the cities while exploring a path for a shortest possible tour. In other embodiments, the different distances between cities can be accounted for by using time.

Multiplying Numbers with an Active Element Machine

This section shows how to multiply numbers with an active element machine. Elements Y0, Y1, Y2, Y3 denote a four bit number and elements Z0, Z1 Z2, Z3 denote a four bit number. The corresponding bit values are y₀, y₁, y₂, y₃ and z₀, z₁, z₂, z₃. The multiplication of y₃ y₂ y₁ y₀*z₃ z₂ z₁ z₀ is shown in Table 13. An active element program is constructed based on Table 13.

The following commands set up elements and connections to perform a four bit multiplication of y₃ y₂ y₁ y₀*z₃ z₂ z₁ z₀ where the result of this multiplication e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀ is stored in elements E7, E6, E5, E4, E3, E2, E1 and E0. The computation is encoded by the elements S_jk corresponds to the product y_(j) z_(k) where S_jk fires if and only if y_(j)=1 and z_(k)=1. The elements C_jk help determine the value of e_(i) represented by element Ei where j+k=

First, two useful program commands are defined.

(Program set_element (Args s xk theta r L) (Element (Time s−2dT) (Name xk) (Threshold theta) (Refractory r) (Last L)) ) (Program set_connection (Args s f xk a w tau) (Connection (Time s−dT) (From f) (To xk) (Amp a) (Width w) (Delay tau)) )

The firing activity of element E₀ expresses the value of e₀. The elements and connections for the product y₀ z₀ which determine the value of e₀ are determined by the following three program commands.

(set_element s E0 3 1 s−2) (set_connection s Y0 E0 2 1 1) (set_connection s Z0 E0 2 1 1)

Table 14 shows the amplitude and threshold used to compute the value of e₀. Table 15 shows the firing patterns for elements S₁₀ and S_(0l) representing the value of products y₁ z₀ and y₀ z₁. Table 16 shows the amplitudes from elements S₁₀ and S_(0l) to elements C_(0l) and C₁₁ and the thresholds of C_(0l) and C₁₁. Table 17 shows the amplitudes from elements C_(0l) and C₁₁ to element E₁ and the threshold of E₁. The firing activity of element E₁ expresses the value of e₁. Below are active element machine commands that express the parameter values of these elements and connections shown in table 14, table 15, table 16 and table 17.

(set_element s S_10 3 1 s−2) (set_connection s Y1 S_10 2 1 1) (set_connection s Z0 S_10 2 1 1) (set_element s S_01 3 1 s−2) (set_connection s Y0 S_01 2 1 1) (set_connection s Z1 S_01 2 1 1) (set_element s C_01 1 1 s−2) (set_element s C_11 3 1 s−2) (set_connection s S_10 C_01 2 1 1) (set_connection s S_10 C_11 2 1 1) (set_connection s S_01 C_01 2 1 1) (set_connection s S_01 C_11 2 1 1) (set_element s E1 1 1 s−2) (set_connection s C_01 E1 2 1 1) (set_connection s C_11 E1 −2 1 1)

Table 18 shows the firing patterns for elements S₂₀, S₁₁, S₀₂ and C₁₁. Table 19 shows the amplitudes from elements S₂₀, S₁₁ S₀₂, C₁₁ to elements C₀₂, C₁₂, C₂₂ C₃₂ and the thresholds of C₀₂, C₁₂, C₂₂ and C₃₂. Table 20 shows the amplitudes from elements C₀₂, C₁₂, C₂₂, C₃₂ to elements P₀₂, P₁₂, P₂₂ and the thresholds of elements P₀₂, P₁₂, P₂₂. Table 21 shows the amplitude and threshold used to compute the value of e₂. The firing activity of element E₂ expresses the value of e₂. Below are active element machine commands that express the parameter values of the elements and connections indicated in table 18, table 19, table 20 and table 21.

(set_element s S_20 3 1 s−2) (set_connection s Y2 S_20 2 1 1) (set_connection s Z0 S_20 2 1 1) (set_element s S_11 3 1 s−2) (set_connection s Y1 S_11 2 1 1) (set_connection s Z1 S_11 2 1 1) (set_element s S_02 3 1 s−2) (set_connection s Y0 S_02 2 1 1) (set_connection s Z2 S_02 2 1 1) (set_element s C_02 1 1 s−2) (set_element s C_12 3 1 s−2) (set_element s C_22 5 1 s−2) (set_element s C_32 7 1 s−2) (set_connection s S_20 C_02 2 1 1) (set_connection s S_20 C_12 2 1 1) (set_connection s S_20 C_22 2 1 1) (set_connection s S_20 C_32 2 1 1) (set_connection s S_11 C_02 2 1 1) (set_connection s S_11 C_12 2 1 1) (set_connection s S_11 C_22 2 1 1) (set_connection s S_11 C_32 2 1 1) (set_connection s S_02 C_02 2 1 1) (set_connection s S_02 C_12 2 1 1) (set_connection s S_02 C_22 2 1 1) (set_connection s S_02 C_32 2 1 1) (set_connection s C_11 C_02 2 1 1) (set_connection s C_11 C_12 2 1 1) (set_connection s C_11 C_22 2 1 1) (set_connection s C_11 C_32 2 1 1) (set_element s P_02 1 1 s−2) (set_element s P_12 1 1 s−2) (set_element s P_22 7 1 s−2) (set_connection s C_02 P_02 2 1 1) (set_connection s C_02 P_12 0 1 1) (set_connection s C_02 P_22 2 1 1) (set_connection s C_12 P_02 −2 1 1) (set_connection s C_12 P_12 2 1 1) (set_connection s C_12 P_22 2 1 1) (set_connection s C_22 P_02 2 1 1) (set_connection s C_22 P_12 2 1 1) (set_connection s C_22 P_22 2 1 1) (set_connection s C_32 P_02 −2 1 1) (set_connection s C_32 P_12 −4 1 1) (set_connection s C_32 P_22 2 1 1) (set_element s E2 1 1 s−2) (set_connection s P_02 E2 2 1 1)

Table 22 shows the firing patterns for elements S₃₀, S₂₁, S₁₂, S₀₃, P₁₂ representing the value of products y₃ z₀, y₂ z₁, y₁ z₂ and y₀ z₃ and the carry value. Table 23 shows the amplitudes from elements S₃₀, S₂₁, S₁₂ and S₀₃ to elements C₀₃, C₁₃, C₂₃, C₃₃, and C₄₃. Table 24 shows the amplitudes from elements C₀₃, C₁₃, C₂₃, C₃₃, and C₄₃ to elements P₀₃, P₁₃, P₂₃ and the thresholds of elements P₀₃, P₁₃, P₂₃. Table 25 shows the amplitude and threshold used to compute the value of e₃. The firing activity of element E₃ expresses the value of e₃. Below are active element machine commands that express the parameter values shown in table 22, table 23, table 24 and table 25.

(set_element s S_30 3 1 s−2) (set_connection s Y3 S_30 2 1 1) (set_connection s Z0 S_30 2 1 1) (set_element s S_21 3 1 s−2) (set_connection s Y2 S_21 2 1 1) (set_connection s Z1 S_21 2 1 1) (set_element s S_12 3 1 s−2) (set_connection s Y1 S_12 2 1 1) (set_connection s Z2 S_12 2 1 1) (set_element s S_03 3 1 s−2) (set_connection s Y0 S_03 2 1 1) (set_connection s Z3 S_03 2 1 1) (set_element s C_03 1 1 s−2) (set_element s C_13 3 1 s−2) (set_element s C_23 5 1 s−2) (set_element s C_33 7 1 s−2) (set_element s C_43 9 1 s−2) (set_connection s S_30 C_03 2 1 1) (set_connection s S_30 C_13 2 1 1) (set_connection s S_30 C_23 2 1 1) (set_connection s S_30 C_33 2 1 1) (set_connection s S_30 C_43 2 1 1) (set_connection s S_21 C_03 2 1 1) (set_connection s S_21 C_13 2 1 1) (set_connection s S_21 C_23 2 1 1) (set_connection s S_21 C_33 2 1 1) (set_connection s S_21 C_43 2 1 1) (set_connection s S_12 C_03 2 1 1) (set_connection s S_12 C_13 2 1 1) (set_connection s S_12 C_23 2 1 1) (set_connection s S_12 C_33 2 1 1) (set_connection s S_12 C_43 2 1 1) (set_connection s S_03 C_03 2 1 1) (set_connection s S_03 C_13 2 1 1) (set_connection s S_03 C_23 2 1 1) (set_connection s S_03 C_33 2 1 1) (set_connection s S_03 C_43 2 1 1) (set_connection s P_12 C_03 2 1 1) (set_connection s P_12 C_13 2 1 1) (set_connection s P_12 C_23 2 1 1) (set_connection s P_12 C_33 2 1 1) (set_connection s P_12 C_43 2 1 1) (set_element s P_03 1 1 s−2) (set_element s P_13 1 1 s−2) (set_element s P_23 7 1 s−2) (set_connection s C_03 P_03 2 1 1) (set_connection s C_03 P_13 0 1 1) (set_connection s C_03 P_23 2 1 1) (set_connection s C_13 P_03 −2 1 1) (set_connection s C_13 P_13 2 1 1) (set_connection s C_13 P_23 2 1 1) (set_connection s C_23 P_03 2 1 1) (set_connection s C_23 P_13 2 1 1) (set_connection s C_23 P_23 2 1 1) (set_connection s C_33 P_03 −2 1 1) (set_connection s C_33 P_13 −4 1 1) (set_connection s C_33 P_23 2 1 1) (set_connection s C_43 P_03 2 1 1) (set_connection s C_43 P_13 0 1 1) (set_connection s C_43 P_23 2 1 1) (set_element s E3 1 1 s−2) (set_connection s P_03 E3 2 1 1)

Table 26 shows the firing patterns for elements S₃₁, S₂₂, S₁₃, P₁₃, P₂₂. Table 27 shows the amplitudes from elements S₃₁, S₂₂, S₁₃, P₁₃, P₂₂ to elements C₀₄, C₁₄, C₂₄, C₃₄, C₄₄ and the thresholds of C₀₄, C₁₄, C₂₄, C₃₄ and C₄₄. Table 28 shows the amplitudes from elements C₀₄, C₁₄, C₂₄, C₃₄, and C₄₄ to elements P₀₄, P₁₄, P₂₄ and the thresholds of elements P₀₄, P₁₄, P₂₄. Table 29 shows the amplitude and threshold used to compute the value of e₄. The firing activity of element E₄ expresses the value of e₄. Below are active element machine commands that express the parameter values shown in table 26, table 27, table 28 and table 29.

(set_element s S_31 3 1 s−2) (set_connection s Y3 S_31 2 1 1) (set_connection s Z1 S_31 2 1 1) (set_element s S_22 3 1 s−2) (set_connection s Y2 S_22 2 1 1) (set_connection s Z2 S_22 2 1 1) (set_element s S_13 3 1 s−2) (set_connection s Y1 S_13 2 1 1) (set_connection s Z2 S_13 2 1 1) (set_element s C_04 1 1 s−2 ) (set_element s C_14 3 1 s−2 ) (set_element s C_24 5 1 s−2 ) (set_element s C_34 7 1 s−2 ) (set_element s C_44 9 1 s−2 ) (set_connection s S_31 C_04 2 1 1) (set_connection s S_31 C_14 2 1 1) (set_connection s S_31 C_24 2 1 1) (set_connection s S_31 C_34 2 1 1) (set_connection s S_31 C_44 2 1 1) (set_connection s S_22 C_04 2 1 1) (set_connection s S_22 C_14 2 1 1) (set_connection s S_22 C_24 2 1 1) (set_connection s S_22 C_34 2 1 1) (set_connection s S_22 C_44 2 1 1) (set_connection s S_13 C_04 2 1 1) (set_connection s S_13 C_14 2 1 1) (set_connection s S_13 C_24 2 1 1) (set_connection s S_13 C_34 2 1 1) (set_connection s S_13 C_44 2 1 1) (set_connection s P_22 C_04 2 1 1) (set_connection s P_22 C_14 2 1 1) (set_connection s P_22 C_24 2 1 1) (set_connection s P_22 C_34 2 1 1) (set_connection s P_22 C_44 2 1 1) (set_connection s P_13 C_04 2 1 1) (set_connection s P_13 C_14 2 1 1) (set_connection s P_13 C_24 2 1 1) (set_connection s P_13 C_34 2 1 1) (set_connection s P_13 C_44 2 1 1) (set_element s P_04 1 1 s−2) (set_element s P_14 1 1 s−2) (set_element s P_24 7 1 s−2) (set_connection s C_04 P_04 2 1 1) (set_connection s C_04 P_14 0 1 1) (set_connection s C_04 P_24 2 1 1) (set_connection s C_14 P_04 −2 1 1) (set_connection s C_14 P_14 2 1 1) (set_connection s C_14 P_24 2 1 1) (set_connection s C_24 P_04 2 1 1) (set_connection s C_24 P_14 2 1 1) (set_connection s C_24 P_24 2 1 1) (set_connection s C_34 P_04 −2 1 1) (set_connection s C_34 P_14 −4 1 1) (set_connection s C_34 P_24 2 1 1) (set_connection s C_44 P_04 2 1 1) (set_connection s C_44 P_14 0 1 1) (set_connection s C_44 P_24 2 1 1) (set_element s E4 1 1 s−2) (set_connection s P_04 E4 2 1 1)

Table 30 shows the firing patterns for elements S₃₂, S₂₃, P₁₄, P₂₃. Table 31 shows the amplitudes from elements S₃₂, S₂₃, P₁₄, P₂₃ to elements C₀₅, C₁₅, C₂₅, C₃₅ and the thresholds of C₀₅, C₁₅, C₂₅, C₃₅. Table 32 shows the amplitudes from elements C₀₅, C₁₅, C₂₅, C₃₅ to elements P₀₅, P₁₅, P₂₅ and the thresholds of elements P₀₅, P₁₅, P₂₅. Table 33 shows the amplitude and threshold used to compute the value of e₅. The firing activity of element E₅ expresses the value of e₅. Below are active element machine commands that express the parameter values shown in table 30, table 31, table 32 and table 33.

(set_element s S_32 3 1 s−2) (set_connection s Y3 S_32 2 1 1) (set_connection s Z2 S_32 2 1 1) (set_element s S_23 3 1 s−2) (set_connection s Y2 S_23 2 1 1) (set_connection s Z3 S_23 2 1 1) (set_element s C_05 1 1 s−2) (set_element s C_15 3 1 s−2) (set_element s C_25 5 1 s−2) (set_element s C_35 7 1 s−2) (set_connection s S_32 C_05 2 1 1) (set_connection s S_32 C_15 2 1 1) (set_connection s S_32 C_25 2 1 1) (set_connection s S_32 C_35 2 1 1) (set_connection s S_23 C_05 2 1 1) (set_connection s S_23 C_15 2 1 1) (set_connection s S_23 C_25 2 1 1) (set_connection s S_23 C_35 2 1 1) (set_connection s P_14 C_05 2 1 1) (set_connection s P_14 C_15 2 1 1) (set_connection s P_14 C_25 2 1 1) (set_connection s P_14 C_35 2 1 1) (set_connection s P_23 C_05 2 1 1) (set_connection s P_23 C_15 2 1 1) (set_connection s P_23 C_25 2 1 1) (set_connection s P_23 C_35 2 1 1) (set_element s P_05 1 1 s−2) (set_element s P_15 1 1 s−2) (set_element s P_25 7 1 s−2) (set_connection s C_05 P_05 2 1 1) (set_connection s C_05 P_15 0 1 1) (set_connection s C_05 P_25 2 1 1) (set_connection s C_15 P_05 −2 1 1) (set_connection s C_15 P_15 2 1 1) (set_connection s C_15 P_25 2 1 1) (set_connection s C_25 P_05 2 1 1) (set_connection s C_25 P_15 2 1 1) (set_connection s C_25 P_25 2 1 1) (set_connection s C_35 P_05 −2 1 1) (set_connection s C_35 P_15 −4 1 1) (set_connection s C_35 P_25 2 1 1) (set_element s E5 1 1 s−2) (set_connection s P_05 E5 2 1 1)

Table 34 shows the firing patterns for elements S₃₃, P₁₅₉ P₂₄. Table 35 shows the amplitudes from elements S₃₃, P₁₅, P₂₄ to elements C₀₆, C₁₆, C₂₆ and the thresholds of C₀₆, C₁₆, C₂₆. Table 36 shows the amplitudes from elements C₀₆, C₁₆, C₂₆ to elements P₀₆, P₁₆ and the thresholds of elements P₀₆, P₁₆. Table 37 shows the amplitude of the connection from element P₀₆ to element E₆ and the threshold of E₆. The firing activity of E₆ expresses the value of e₆. Below are active element machine commands that express the parameter values shown in table 34, table 35, table 36 and table 37.

(set_element s S_33 3 1 s−2) (set_connection s Y3 S_33 2 1 1) (set_connection s Z3 S_33 2 1 1) (set_element s C_06 1 1 s−2) (set_element s C_16 3 1 s−2) (set_element s C_26 5 1 s−2) (set_connection s S_33 C_06 2 1 1) (set_connection s S_33 C_16 2 1 1) (set_connection s S_33 C_26 2 1 1) (set_connection s P_15 C_06 2 1 1) (set_connection s P_15 C_16 2 1 1) (set_connection s P_15 C_26 2 1 1) (set_connection s P_24 C_06 2 1 1) (set_connection s P_24 C_16 2 1 1) (set_connection s P_24 C_26 2 1 1) (set_element s P_06 1 1 s−2 ) (set_element s P_16 1 1 s−2 ) (set_connection s C_06 P_06 2 1 1) (set_connection s C_06 P_16 0 1 1) (set_connection s C_16 P_06 −2 1 1) (set_connection s C_16 P_16 2 1 1) (set_connection s C_26 P_06 2 1 1) (set_connection s C_26 P_16 2 1 1) (set_element s E6 1 1 s−2) (set_connection s P_06 E6 2 1 1)

The firing activity of element E₇ represents bit e₇. When element P₁₆ is firing, this means that there is a carry so E₇ should fire. The following commands accomplish this.

(set_element s E7 1 1 s−2) (set_connection s P_16 E7 2 1 1)

Table 38 shows how the active element machine commands were designed to compute 1110*0111. Suppose that the AEM commands from the previous sections are called with s=2. Then y₀=0. y₁=1. y₂=1. y₃=1. z₀=1. z₁=1. z₂=1. z₃=0. Thus, fire input elements Y₁, Y₂ and Y₃ at time 2. Similarly, fire input elements Z₀, Z₁ and Z₂ at time 2.

(set_element 2 Y1 1 1 0) (set_connection 2 Y1 Y1 2 1+dT 1) (fire (Time 2) (Name Y1)) (set_element 2 Y2 1 1 0) (set_connection 2 Y2 Y2 2 1+dT 1) (fire (Time 2) (Name Y2)) (set_element 2 Y3 1 1 0) (set_connection 2 Y3 Y3 2 1+dT 1) (fire (Time 2) (Name Y3)) (set_element 2 Z0 1 1 0) (set_connection 2 Z0 Z0 2 1+dT 1) (fire (Time 2) (Name Z0)) (set_element 2 Z1 1 1 0) (set_connection 2 Z1 Z1 2 1+dT 1) (fire (Time 2) (Name Z1)) (set_element 2 Z2 1 1 0) (set_connection 2 Z2 Z2 2 1+dT 1) (fire (Time 2) (Name Z2))

Element E₀ never fires because E₀ only receives a pulse of amplitude 2 from Z₀ and has threshold 3. The fact that E₀ never fires represents that e₀=0.

In regard to the value of e₁, element S₁₀ fires at time 3 because Y₁ and Z₀ fire at time 2 and S₁₀ has a threshold of 3 and receives a pulse of amplitude 2 from Y₁, and Z₀. The following commands set these values.

(set_element s S_10 3 1 s−2) (set_connection s Y1 S_10 2 1 1) (set_connection s Z0 S_10 2 1 1)

Element S_(0l) does not fire at time 3 because it only receives a pulse of amplitude 2 from element Z₁ and has threshold 3. The firing of S₁₀ at time 3 causes C₀₁ to fire at time 4 because C₀₁'s threshold is 1. The following commands set up these element and connection values.

(set_element s C_01 1 1 s−2) (set_connection s S_10 C_01 2 1 1)

The commands

(set_element s E1 1 1 s−2) (set_connection s C_01 E1 2 1 1) cause E₁ to fire at time 5 and E₁ continues to fire indefinitely because the input elements Y₁, Y₂, Y₃, Z₀, Z₁ and Z₂ continue to fire at time steps 3, 4, 5, 6, 7, 8, . . . . The firing of element E₁ indicates that e₁=1.

In regard to the value of e₂, since elements Y₁ and Z₁ fire at time 2, element S₁₁ fires at time 3. Since elements Y₂ and Z₀ fire at time 2, element S₂₀ also fires at time 3. From the following commands

(set_element 2 C_02 1 1 0) (set_element 2 C_12 3 1 0) (set_connection 2 S_20 C_02 2 1 1) (set_connection 2 S_20 C_12 2 1 1) (set_connection 2 S_11 C_02 2 1 1) (set_connection 2 S_11 C_12 2 1 1) then elements C₀₂ and C₁₂ fire at time 4.

Element P₁₂ fires at time 5 because of the commands

(set_element 2 P_12 1 1 0) (set_connection 2 C_12 P_12 2 1 1) (set_connection 2 C_02 P_12 0 1 1)

Observe that element P₀₂ does not fire because C₁₂ sends a pulse with amplitude −2 and C₀₂ sends a pulse with amplitude 2 and element P₀₂ has threshold 1 as a consequence of command

(set_element 2 P_(—)02 1 1 0).

Since P₀₂ does not fire, element E₂ does not fire as it threshold is 1 and the only connection to element E₂ is from P₀₂: (set_connection 2 P_(—)02 E2 2 1 1). Since element E₂ does not fire, this indicates that e₂=0.

In regard to the value of e₃, since elements Y₃ and Z₀ fire at time 2, element S₃₀ fires at time 3. Since elements Y₂ and Z₁ fire at time 2, element S₂₁ fires at time 3. Since elements Y₁ and Z₂ fire at time 2, element S₁₂ fires at time 3. S₀₃ does not fire. From the following commands

(set_element 2 C_03 1 1 0) (set_element 2 C_13 3 1 0) (set_element 2 C_23 5 1 0) (set_element 2 C_33 7 1 0) (set_connection 2 S_30 C_03 2 1 1) (set_connection 2 S_30 C_13 2 1 1) (set_connection 2 S_30 C_23 2 1 1) (set_connection 2 S_30 C_33 2 1 1) (set_connection 2 S_21 C_03 2 1 1) (set_connection 2 S_21 C_13 2 1 1) (set_connection 2 S_21 C_23 2 1 1) (set_connection 2 S_21 C_33 2 1 1) (set_connection 2 S_12 C_03 2 1 1) (set_connection 2 S_12 C_13 2 1 1) (set_connection 2 S_12 C_23 2 1 1) (set_connection 2 S_12 C_33 2 1 1) (set_connection 2 P_12 C_03 2 1 1) (set_connection 2 P_12 C_13 2 1 1) (set_connection 2 P_12 C_23 2 1 1) (set_connection 2 P_12 C_33 2 1 1) then elements C₀₃, C₁₃, C₂₃ fire at time 4 and they will continue to fire every time step 5, 6, 7, 8 . . . because the elements Y₁, Y₂, Y₃, Z₀, Z₁ and Z₂ continue to fire at time steps 3, 4, 5, 6, 7, 8, . . . .

As a result of P₁₂ firing at time 5, C₃₃ fires at time 6, so at time 7, only P₂₃ fires. As a result, the long term behavior (after time step 7) of P₀₃ does not fire. Thus, E₃ does not fire after time step 7, which indicates that e₃=0.

Similar to that of element E₃, in the long term element E₄ does not fire, which indicates that e₄=0. Similarly, in the long term element E₅ fires, which indicates that e₅=1. Similarly, in the long term element E₆ fires, which indicates that e₆=1. Similarly, in the long term element E₇ does not fire, which indicates that e₇=0.

As a consequence, multiplication of 1110*0111 equals 1100010 in binary, which represents that 14*7=98 in base 10. This active element program can execute any multiplication of two four bit binary numbers. Similar to the multiplication just described, tables 39, 40 and 41 show the multiplication steps for 11*9=99; 15*14=210; and 15*15=225.

AEM Firing Patterns Execute a Digital Computer Program

In some embodiments, an AEM using randomness deterministically executes a universal Turing machine or a von Neumann machine (digital computer program). In some embodiments, the randomness is generated using quantum phenomena. In some embodiments, the randomness is generated from a physical process. In some embodiments, the firing patterns of the active elements computing the execution of these machines are Turing incomputable. In some embodiments, the AEM accomplishes this by executing a universal Turing machine or von Neumann machine instructions with random firing interpretations. In some embodiments, if the state and tape contents of the universal Turing machine—represented by the AEM elements and connections—and the random bits generated from the random—in some embodiments, quantum—source are kept perfectly secret and no information is leaked about the dynamic connections in the AEM, then it is Turing incomputable to construct a translator Turing machine that maps the random firing interpretations back to the sequence of instructions executed by the universal Turing machine or von Neumann machine. As a consequence, in some embodiments, the AEM can deterministically execute any Turing machine (digital computer program) with active element firing patterns that are Turing incomputable. Since Turing incomputable AEM firing behavior can deterministically execute a universal Turing machine or digital computer with a finite active element machine using quantum randomness, this creates a novel computational procedure ([6], [32]). In [20], Lewis and Papadimitriou discuss the prior art notion of a digital computer's computational procedure:

Because the Turing machines can carry out any computation that can be carried out by any similar type of automata, and because these automata seem to capture the essential features of real computing machines, we take the Turing machine to be a precise formal equivalent of the intuitive notion of algorithm: nothing will be considered as an algorithm if it cannot be rendered as a Turing machine.

The principle that Turing machines are formal versions of algorithms and that no computational procedure will be considered as an algorithm unless it can be presented as a Turing machine is known as Church's thesis or the Church-Turing Thesis. It is a thesis, not a theorem, because it is not a mathematical result: It simply asserts that a certain informal concept corresponds to a certain mathematical object. It is theoretically possible, however, that Church's thesis could be overthrown at some future date, if someone were to propose an alternative model of computation that was publicly acceptable as fulfilling the requirement of finite labor at each step and yet was provably capable of carrying out computations that cannot be carried out by any Turing machine. No one considers this likely.

In a cryptographic system, Shannon [28] defines the notion of perfect secrecy.

Perfect Secrecy is defined by requiring of a system that after a cryptogram is intercepted by the enemy the a posteriori probabilities of this cryptogram representing various messages be identically the same as the a priori probabilities of the same messages before the interception.

In this context, perfect secrecy means that no information is ever released or leaked about the state and the contents of the universal Turing machine tape, the random bits generated from a quantum source and the dynamic connections of the active element machine.

In [19], Kocher et al. present differential power analysis. Differential power analysis obtains information about cryptographic computations executed by register machine hardware, by statistically analyzing the electromagnetic radiation leaked by the hardware during its computation. In some embodiments, when a quantum active element computing system is built so that its internal components remain perfectly secret or close to perfectly secret, then it may be extremely challenging for an adversary to carry out types of attacks such as differential power analysis.

Active Element Machine Interpretations of Boolean Functions

In this section, the same boolean function is computed by two or more distinct active element firing patterns, which can be executed at distinct times or by different circuits (two or more different parts) in the active element machine. These methods provide useful embodiments in a number of ways. They show how digital computer program computations can be computed differently at distinct instances. In some embodiments, distinct instances are two or more different times. In some embodiments, distinct instances use different elements and connections of the active element machine to differently compute the same Boolean function. The methods shown here demonstrate the use of level sets so that multiple active element machine firing patterns may compute the same boolean function or computer program instruction. Third, these methods demonstrate the utility of using multiple, dynamic firing interpretations to perform the same task—for example, execute a computer program—or represent the same knowledge.

The embodiments shown here enable one or more digital computer program instructions to be computed differently at different instances. In some embodiments, these different instances are different times. In some embodiments, these different instances of computing the program instruction are executed by different collections of active elements and connections in the active element machine. In some embodiments, the computer program may be an active element machine program.

The following procedure uses a random physical process to either fire input element I or not fire I at time t=n where n is a natural number {0, 1, 2, 3, . . . }. This random sequence of 0 and 1's can be generated by quantum optics, or quantum effects in a semiconductor material or other physical phenomena. The procedure is used to help execute the same computation with multiple interpretations. In some embodiments, this same computation is a program instruction executed at two different instances.

Procedure 1. Randomness generates an AEM, representing a real number in [0, 1]. Using a random process to fire or not fire one input element I at each unit of time, a finite active element program can represent a randomly generated real number in the unit interval [0, 1]. In some embodiments, the random process is physically contained in the active element machine. In other embodiments, the random process is outside of the active element machine.

The Meta command and a random sequence of bits creates active elements 0, 1, 2, . . . that store the binary representation b₀ b₁ b₂ . . . of real number xε[0, 1]. If input element I fires at time t=n, then b_(n)=1; thus, create active element n so that after t=n, element n fires every unit of time indefinitely. If input element I doesn't fire at time t=n, then b_(n)=0 and active element n is created so that it never fires. The following finite active element machine program exhibits this behavior.

(Program C (Args t) (Connection (Time t) (From I) (To t) (Amp 2) (Width 1) (Delay 1)) (Connection (Time t+1+dT) (From I) (To t) (Amp 0)) (Connection (Time t) (From t) (To t) (Amp 2) (Width 1) (Delay 1)) ) (Element (Time clock) (Name clock) (Threshold 1) (Refractory 1) (Last −1)) (Meta (Name I) (C (Args clock)))

Suppose a sequence of random bits—obtained from the environment or from a physical process inside the active element machine—begins with 1, 0, 1, . . . . Thus, input element I fires at times 0, 2, . . . . At time 0, the following commands are executed.

(Element (Time 0) (Name 0) (Threshold 1) (Refractory 1) (Last −1)) (C (Args 0))

The execution of (C (Args 0)) causes three connection commands to execute.

(Connection (Time 0) (From I) (To 0) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 1+dT) (From I) (To 0) (Amp 0)) (Connection (Time 0) (From 0) (To 0) (Amp 2) (Width 1) (Delay 1))

Because of the first connection command

(Connection (Time 0) (From I) (To 0) (Amp 2) (Width 1) (Delay 1))

the firing of input element I at time 0 sends a pulse with amplitude 2 to element 0. Thus, element 0 fires at time 1. Then at time 1+dT, a moment after time 1, the connection from input element I to element 0 is removed. At time 0, a connection from element 0 to itself with amplitude 2 is created. As a result, element 0 continues to fire indefinitely, representing that b₀=1. At time 1, command (Element (Time 1) (Name 1) (Threshold 1) (Refractory 1) (Last −1)) is created. Since element 1 has no connections into it and threshold 1, element 1 never fires. Thus b₁=0. At time 2, input element I fires, so the following commands are executed.

(Element (Time 2) (Name 2) (Threshold 1) (Refractory 1) (Last −1)) (C (Args 2))

The execution of (C (Args 2)) causes the three connection commands to execute.

(Connection (Time 2) (From I) (To 2) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 3+dT) (From I) (To 2) (Amp 0)) (Connection (Time 2) (From 2) (To 2) (Amp 2) (Width 1) (Delay 1))

Because of the first connection command

(Connection (Time 2) (From I) (To 2) (Amp 2) (Width 1) (Delay 1))

the firing of input element I at time 2 sends a pulse with amplitude 2 to element 2. Thus, element 2 fires at time 3. Then at time 3+dT, a moment after time 3, the connection from input element I to element 2 is removed. At time 2, a connection from element 2 to itself with amplitude 2 is created. As a result, element 2 continues to fire indefinitely, representing that b₂=1.

Active Element Machine Firing Patterns

During a window of time, firing patterns can be put in 1-to-1 correspondence with the boolean functions ƒ:{0, 1}^(n)→{0, 1}. In the next section, the firing pattern methods explained here are combined with procedure 1 so that a randomly chosen firing pattern can compute the functions used to execute a universal Turing machine. Consider four active elements X₀, X₁, X₂ and X₃ during window of time W=[a, b]. The refractory periods of X₀, X₁, X₂ and X₃ are chosen so that each X_(k) either fires or doesn't fire during window W. Thus, there are sixteen distinct firing patterns. Five of these firing patterns are shown in FIGS. 1, 2, 3, 4, and 5.

A one-to-one correspondence is constructed with the sixteen boolean functions of the form ƒ: {0, 1}×{0, 1}→{0, 1}. These boolean functions comprise the binary operators: and

, or v, nand

, equal

, and so on. One of these firing patterns is distinguished from the other fifteen by building the appropriate connections to element P, which in the general case represents the output of a boolean function ƒ: {0, 1}^(n)→{0, 1}. A key notion is that element P fires within the window of time W if and only if P receives a unique firing pattern from elements X₀, X₁, X₂ and X₃. (This is analogous to the notion of the grandmother nerve cell that only fires if you just saw your grandmother.) The following definition covers the Boolean interpretation explained here and also handles more complex types of interpretations.

Definition 2.1. Number of Firings During a Window

Let X denote the set of active elements {X₀, X₁, . . . , X_(n-1)} that determine the firing pattern during the window of time W. Then |(X_(k), W)|=the number of times that element X_(k) fired during window of time W. Thus, define the number of firings during window W as

${\left( {X,W} \right)} = {\sum\limits_{k = 0}^{n - 1}\; {{\left( {X_{k},W} \right)}.}}$

Observe that |(X, W)|=0 for firing pattern 0000 shown in FIG. 1 and |(X, W)|=2 for firing pattern 0011. To isolate a firing pattern so that element P only fires if this unique firing pattern occurs, set the threshold of element P=2|(X, W)|−1.

The element command for P is:

(Element (Time a−dT) (Name P) (Threshold 2|(X, W)|−1) (Refractory b−a) (Last 2a−b))

Further, if element X_(k) doesn't fire during window of time W, then set the amplitude of the connection from X_(k) to P to −2. If element X_(k) does fire during window of time W, then set the amplitude of the connection from X_(k) to P equal to 2. For each element X_(k), the pulse width is set to |W|=b−a. Each connection from X_(k) to P is based on whether X_(k) is supposed to fire or is not supposed to fire during W. If X_(k) is supposed to fire during W, then the following connection is established. (Connection (Time a−dT) (From X_k) (To P) (Amp 2) (Width b−a) (Delay 1)) If X_(k) is not supposed to fire during W, then the following connection is established.

(Connection (Time a−dT) (From X_k) (To P) (Amp −2) (Width b−a) (Delay 1))

The firing pattern is already known because it is determined based on a random source of bits received by input elements, as discussed in procedure 1. Consequently, −2|(X, W)| is already known. How an active element circuit is designed to create a firing pattern that computes the appropriate boolean function is discussed in the following example.

Example. Computing ⊕ with Firing Pattern 0010

Consider firing pattern 0010. In other words, X₂ fires but the other elements do not fire. The AEM is supposed to compute the boolean function exclusive-or A⊕B=(A

B)

(

A

B). The goal here is to design an AEM circuit such that A⊕B=1 if and only if the firing pattern for X₀, X₁, X₂, X₃ is 0010. Following definition 2.1, as a result of the distinct firing pattern during W, if A⊕B=1 then P fires. If A⊕B=0 then P doesn't fire. Below are the commands that connect elements A and B to elements X₀, X₁, X₂, X₃.

(Connection (Time a−2) (From A) (To X_0) (Amp 2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_0) (Amp 2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_0) (Threshold 3) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_1) (Amp −2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_1) (Amp −2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_1) (Threshold −1) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_2) (Amp 2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_2) (Amp 2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_2) (Threshold 1) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_3) (Amp 2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_3) (Amp 2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_3) (Threshold 3) (Refractory b−a) (Last 2a−b))

There are four cases for A⊕B: 1⊕0, 0⊕1, 1⊕1 and 0⊕0. In regard to this, choose the refractory periods so that A and B either fire or don't fire at t=0. Recall that W=[a, b]. In this example, assume a=2 and b=3. Thus, all refractory periods of X₀, X₁, X₂, X₃ are 1 and all last time fireds are 1. All pulse widths are the length of the window W+1 which equals 2.

Case 1. Element A fires at time t=0 and element B doesn't fire at t=0.

Element X₀ receives a pulse from A with amplitude 2 at time t=2. Element X₀ doesn't fire because its threshold=3. Element X₁ receives a pulse from A with amplitude −2 at time t=2. Element X₁ doesn't fire during W because X₁ has threshold=−1. Element X₂ receives a pulse from A with amplitude 2. Element X₂ fires at time t=2 because its threshold is 1. Element X₃ receives a pulse from A with amplitude 2 but doesn't fire during window W because X₃ has threshold=3.

Case 2. Element X₀ receives a pulse from B with amplitude 2 at time t=2.

Element X₀ doesn't fire because its threshold=3. Element X₁ receives a pulse from B with amplitude −2 at time t=2. Element X₁ doesn't fire during W because X₁ has threshold=−1. Element X₂ receives a pulse from B with amplitude 2. Element X₂ fires at time t=2 because its threshold is 1. Element X₃ receives a pulse from B with amplitude 2, but doesn't fire during window W because X₃ has threshold=3.

Case 3. Element A fires at time t=0 and element B fires at t=0.

Element X₀ receives two pulses from A and B each with amplitude 2 at time t=2. Element X₀ fires because its threshold=3. Element X₁ receives two pulses from A and B each with amplitude −2 at time t=2. Element X₁ doesn't fire during W because X₁ has threshold=−1. Element X₂ receives two pulses from A and B each with amplitude 2. Element X₂ fires at time t=2 because its threshold is 1. Element X₃ receives two pulses from A and B each with amplitude 2. Element X₃ fires at time t=2 because X₃ has threshold=3.

Case 4. Element A doesn't fire at time t=0 and element B doesn't fire at t=0. Thus, elements X₀, X₂ and X₃ do not fire because they have positive thresholds. Element X₁ fires at t=2 because it has threshold=−1.

For the desired firing pattern 0010, the threshold of P=2|(X, W)|−1=1. Below is the element command for P.

(Element (Time 2−dT) (Name P) (Threshold 1) (Refractory 1) (Last 1)).

Below are the connection commands for making P fire if and only if firing pattern 0010 occurs during W.

(Connection (Time 2−dT) (From X_0) (To P) (Amp −2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_1) (To P) (Amp −2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_2) (To P) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_3) (To P) (Amp −2) (Width 1) (Delay 1))

For cases 1 and 2 (i.e., 1⊕0 and 0⊕1) only X₂ fires. A moment before X₂ fires at t=2 (i.e., −dT), the amplitude from X₂ to P is set to 2. At time t=2, a pulse with amplitude 2 is sent from X₂ to P, so P fires at time t=3 since its threshold=1. In other words, 1⊕0=1 or 0⊕1=1 has been computed. For case 3, (1⊕1), X₀, X₂ and X₃ fire. Thus, two pulses each with amplitude=−2 are sent from X₀ and X₃ to P. And one pulse with amplitude 2 is sent from X₂ to P. Thus, P doesn't fire. In other words, 1⊕1=0 has been computed. For case 4, (0⊕0), X₁ fires. One pulse with amplitude=−2 is sent to X₂. Thus, P doesn't fire. In other words, 0⊕0=0 has been computed.

Level Set Separation Rules

This section describes how any of the sixteen boolean functions are uniquely mapped to one of the sixteen firing patterns by an appropriate active element machine program. The domain {0, 1}×{0, 1} of the sixteen boolean functions has four members {(0, 0), (1, 0), (0, 1), (1, 1)}. Furthermore, for each active element X_(k), separate these members based on the (amplitude from A to X_(k), amplitude from B to X_(k), threshold of X_(k), element X_(k)) quadruplet. For example, the quadruplet (0, 2, 1, X₁) separates {(1, 1), (0, 1)} from {(1, 0), (0, 0)} with respect to X₁. Recall that A=1 means A fires and B=1 means B fires. Then X₁ will fire with inputs {(1, 1), (0, 1)} and X₁ will not fire with inputs {(1, 0), (0, 0)}. The separation rule is expressed as

$\left. \left( {0,2,1,X_{1}} \right)\leftrightarrow{\frac{\left\{ {\left( {1,1} \right),\left( {0,1} \right)} \right\}}{\left\{ {\left( {1,0} \right),\left( {0,0} \right)} \right\}}.} \right.$

Similarly,

$\left. \left( {0,{- 2},{- 1},X_{2}} \right)\leftrightarrow\frac{\left\{ {\left( {1,0} \right),\left( {0,0} \right)} \right\}}{\left\{ {\left( {1,1} \right),\left( {0,1} \right)} \right\}} \right.$

indicates that X₂ has threshold −1 and amplitudes 0 and −2 from A and B respectively. Further, X₂ will fire with inputs {(1, 0), (0, 0)} and will not fire with inputs {(1, 1), (0, 1)}.

Table 1 shows how to compute all sixteen boolean functions ƒ_(k): {0, 1}×{0, 1}→{0, 1}. For each X, use one of 14 separation rules to map the level set ƒ_(k) ⁻¹{1} or alternatively map the level set ∂_(k) ⁻¹{0} to one of the sixteen firing patterns represented by X₀, X₁, X₂ and X₃. The level set method works as follows.

Suppose the nand boolean function ƒ₁₃=

(A

B) is to be computed with the firing pattern 0101. Observe that ƒ₁₃ ⁻¹{1}={(1, 0), (0, 1), (0, 0)}. Thus, the separation rules

$\left. \left( {2,2,3,X_{k}} \right)\leftrightarrow\frac{\left\{ \left( {1,1} \right) \right\}}{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {0,0} \right)} \right\}} \right.$

for k in {0, 2} work because X₀ and X₂ fire if and only if A fires and B fires. Similarly,

$\left. \left( {{- 2},{- 2},{- 3},X_{j}} \right)\leftrightarrow\frac{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {0,0} \right)} \right\}}{\left\{ \left( {1,1} \right) \right\}} \right.$

for j in {1, 3} work because X₁ and X₃ don't fire if and only if A fires and B fires. These rules generate the commands.

(Connection (Time a−2) (From A) (To X_0) (Amp 2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_0) (Amp 2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_0) (Threshold 3) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_1) (Amp −2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_1) (Amp −2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_1) (Threshold −3) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_2) (Amp 2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_2) (Amp 2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_2) (Threshold 3) (Refractory b−a) (Last 2a−b)) (Connection (Time a−2) (From A) (To X_3) (Amp −2) (Width b−a+1) (Delay 2)) (Connection (Time a−2) (From B) (To X_3) (Amp −2) (Width b−a+1) (Delay 2)) (Element (Time a−2) (Name X_3) (Threshold −3) (Refractory b−a) (Last 2a−b))

The five commands make element P fire if and only if firing pattern 0101 occurs.

(Element (Time 2−dT) (Name P) (Threshold 3) (Refractory 1) (Last 1)) (Connection (Time 2−dT) (From X_0) (To P) (Amp −2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_1) (To P) (Amp 2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_2) (To P) (Amp −2) (Width 1) (Delay 1)) (Connection (Time 2−dT) (From X_3) (To P) (Amp 2) (Width 1) (Delay 1))

Case 1:

(0

0). A doesn't fire and B doesn't fire. Thus, no pulses reach X₁ and X₃, who each have threshold −3. Thus, X₁ and X₃ fire. Similarly, no pulses reach X₀ and X₂, who each have threshold 3. Thus, the firing pattern 0101 shown in table 10 causes P to fire because element X₁ and X₃ each send a pulse with amplitude 2 to P which has threshold 3. Therefore,

(0

0)=1 is computed.

Case 2:

(1

0). A fires and B doesn't fire. Thus, one pulse from A with amplitude 2 reaches X₀ and X₂, who each have threshold 3. Thus, X₀ and X₂ don't fire. Similarly, one pulse from A with amplitude −2 reaches X₁ and X₃, who each have threshold −3. Thus, the firing pattern 0101 shown in table 11 causes P to fire because element X₁ and X₃ each send a pulse with amplitude 2 to P which has threshold 3. Therefore,

(1

0)=1 is computed.

Case 3:

(0

1). A doesn't fire and B fires. Thus, one pulse from B with amplitude 2 reaches X₀ and X₂, who each have threshold 3. Thus, X₀ and X₂ don't fire. Similarly, one pulse from B with amplitude −2 reaches X₁ and X₃, who each have threshold −3. Thus, the firing pattern 0101 shown in table 12 causes P to fire because element X₁ and X₃ each send a pulse with amplitude 2 to P which has threshold 3. Therefore,

(0

1)=1 is computed.

Case 4:

(1

1). A fires and B fires. Thus, two pulses each with amplitude 2 reach X₀ and X₂, who each have threshold 3. Thus, X₀ and X₂ fire. Similarly, two pulses each with amplitude −2 reach X₁ and X₃, who each have threshold −3. As a result, X₁ and X₃ don't fire. Thus, the firing pattern 1010 shown in table 13 prevents P from firing because X₀ and X₂ each send a pulse with amplitude −2 to P which has threshold 3. Therefore,

(1

1)=0 is computed.

Overall, any one of the sixteen boolean functions in table 1 are uniquely mapped to one of the sixteen firing patterns by an appropriate AEM program. These mappings can be chosen arbitrarily: as a consequence, each register machine instruction can be executed at different times using distinct AEM firing representations.

Executing a Digital Computer with Random Firing Interpretations

A universal Turing Machine (UTM) is a Turing machine that can execute the computation of any Turing Machine by reading the other Turing Machine's description and input from the UTM's tape. Table 2 shows Minsky's universal Turing machine described in [24]. This means that this universal Turing machine can execute any program that a digital computer or a distruted system computers or a vonNeumann machine can execute.

The elements of {0, 1}² are denoted as {00, 01, 10, 11}. Create a one-to-one correspondence between the tape symbols in the alphabet

and the elements in {0, 1}² as follows: 0

00, 1

01, y

10 and A

11. Furthermore, consider the following correspondence of the states with the elements of {0, 1}³: q₁

001, q₂

010, q₃

011, q₄

100, q₅

101, q₆

110, q₇

111 and the halting state h

000. Further consider L

0 and R

1 in {0, 1}. An active element machine is designed to compute the universal Turing Machine program η shown in table 3. Following the methods in the previous section, multiple AEM firing interpretations are created that compute η. When the universal Turing machine halts, η(011, 00)=(000, 00, h), this special case is handled with a halting firing pattern

that the active element machine enters. Concatenate the three boolean variables U, W, X to represent the current state of the UTM. The two boolean variables Y, Z represent the current tape symbol. From table 3, observe that η=(η₀ η₁ η₂, η₃ η₄, η₅). For each k such that 0≦k≦5, the level sets of the function η_(k): {0, 1}³×{0, 1}²→{0, 1} are shown below.

η₀ ⁻¹(UWX, YZ){1}={(111, 10), (111, 01), (111, 00), (110, 11), (110, 10), (110, 01), (101, 11), (101, 10), (101, 01), (100, 11), (100, 10), (100, 01), (100, 00), (011, 11) (010, 11)}

η₀ ⁻¹(UWX, YZ){0}={(111, 11), (110, 00), (101, 00), (011, 10), (011, 01), (011, 00), (010, 10), (010, 01), (010, 00), (001, 11), (001, 10), (001, 01), (001, 00), (000, 11), (000, 10), (000, 01), (000, 00)}

η₁ ⁻¹(UWX, YZ){1}={(111, 11), (111, 10), (111, 01), (111, 00), (110, 11), (110, 10), (110, 01), (110, 00), (101, 00), (100, 01), (011, 10), (011, 01), (010, 11), (010, 01), (010, 00), (001, 01)}

η₁ ⁻¹(UWX, YZ){0}={(101, 11), (101, 10), (101, 01), (100, 11), (100, 10), (100, 00), (011, 11), (011, 00), (010, 10), (001, 11), (001, 10), (001, 00), (000, 11), (000, 10), (000, 01), (000, 00)}

η₂ ⁻¹(UWX, YZ){1}={011, 10), (111, 01), (110, 00), (101, 11), (101, 10), (101, 01), (101, 00), (100, 01), (100, 00), (011, 10), (011, 01), (010, 10), (001, 11), (001, 10), (001, 00)}

η₂ ⁻¹(UWX, YZ){0}={(111, 11), (111, 00), (110, 11), (110, 10), (110, 01), (100, 11), (100, 10), (011, 11), (011, 00), (010, 11), (010, 01), (010, 00), (001, 01), (000, 11), (000, 10), (000, 01), (000, 00)}

η₃ ⁻¹(UWX, YZ){1}={(111, 00), (110, 00), (110, 01), (110, 10), (101, 00), (101, 01), (101, 10), (100, 00), (100, 10), (011, 01), (011, 10), (010, 11), (010, 01), (010, 00)}

η₄ ⁻¹(UWX, YZ){0}={(111, 01), (111, 10), (111, 11), (110, 11), (101, 11), (100, 01), (100, 11), (011, 00), (011, 11), (010, 10), (001, 00), (001, 01), (001, 10), (001, 11), (000, 01), (000, 10), (000, 11), (000, 00)}

η₄ ⁻¹(UWX, YZ){1}={(111, 01), (110, 11), (110, 01), (110, 00), (101, 11), (101, 01), (100, 11), (100, 01), (011, 11), (011, 01), (010, 01), (001, 11), (001, 01)}

η₄ ⁻¹(UWX, YZ){0}={(111, 11), (111, 10), (111, 00), (110, 10), (101, 10), (101, 00), (100, 10), (100, 00), (011, 10), (011, 00), (010, 11), (010, 10), (010, 00), (001, 10), (001, 00), (000, 11), (000, 10), (000, 01), (000, 00)}

η₅ ⁻¹(UWX, YZ){1}=(111, 11), (111, 10), (111, 01), (111, 00), (110, 11), (110, 10), (110, 01), (101, 11), (101, 10), (101, 01), (100, 00), (010, 11), (010, 01), (010, 00)

η₅ ⁻¹(UWX, YZ){0}={(110, 00), (101, 00), (100, 11), (100, 10), (100, 01), (011, 11), (011, 10), (011, 01), (010, 10), (001, 11), (001, 10), (001, 01), (001, 00), (000, 11), (000, 10)}(000, 01), (000, 00)

The level set η₅ ⁻¹(UWX, YZ){h}={(011, 00)} is the special case when the universal Turing machine halts. At this time, the active element machine reaches a halting firing pattern H. The next example copies one element's firing state to another element's firing state, which helps assign the value of a random bit to an active element and perform other functions in the UTM.

Copy Program.

This active element program copies active element a's firing state to element b.

(Program copy (Args s t b a) (Element (Time s−1) (Name b) (Threshold 1) (Refractory 1) (Last s−1)) (Connection (Time s−1)(From a) (To b)(Amp 0) (Width 0) (Delay 1)) (Connection (Time s) (From a) (To b) (Amp 2) (Width 1) (Delay 1)) (Connection (Time s) (From b) (To b) (Amp 2) (Width 1) (Delay 1)) (Connection (Time t) (From a) (To b) (Amp 0) (Width 0) (Delay 1)) )

When the copy program is called, active element b fires if a fired during the window of time [s, t). Further, a connection is set up from b to b so that b will keep firing indefinitely. This enables b to store active element a's firing state. The following procedure describes the computation of the Turing program with random firing interpretations.

Procedure 2. Computing Turing Program with Random Firing Patterns Consider function η₃: {0, 1}⁵→0, 1} as previously defined. The following scheme for mapping boolean values 1 and 0 to the firing of an active element is used. If active element U fires during window W, then this corresponds to input U=1 in η₃; if active element U doesn't fire during window W, then this corresponds to input U=0 in η₃. When U fires, W doesn't fire, X fires, Y doesn't fire and Z doesn't fire, this corresponds to computing η₃ (101, 00). The value 1=η₃(101, 00) is the underlined bit in (011, 10, 0), which is located in row 101, column 00 of table 3. Procedure 1 and the separation rules in table 4 are synthesized so that is computed using random active element firing patterns. In other words, the boolean function r₇₃ can be computed using an active element machine's dynamic interpretation. The dynamic part of the interpretation is determined by the random bits received from a quantum source. The firing activity of active element P₃ represents the value of η₃ (UWX, YZ). Fourteen random bits are read from a quantum random generator—for example, see [5]. These random bits are used to create a corresponding random firing pattern of active elements R₀, R₁, . . . R₁₃. Meta commands dynamically build active elements and connections based on the separation rules in table 4 and the firing activity of elements R₀, R₁, . . . R₁₃. These dynamically created active elements and connections determine the firing activity of active element P₃ based on the firing activity of active elements U, W, X, Y and Z. The details of this procedure are described below.

Read fourteen random bits a₀, a₁, . . . and a₁₃ from a quantum source. The values of these random bits are stored in active elements R₀, R₁, . . . R₁₃. If random bit a_(k)=1, then R_(k) fires; if random bit a_(k)=0, then R_(k) doesn't fire.

Set up dynamical connections from active elements U, X, W, Y, Z to elements D₀, D₁, . . . D₁₃. These connections are based on Meta commands that use the firing pattern from elements R₀, R₁, . . . R₁₃.

(Program set_dynamic_C (Args s t f xk a w tau rk)  (Connection (Time s−dT) (From f) (To xk) (Amp −a) (Width w) (Delay  tau))  (Meta (Name rk) (Window s t) (Connection (Time t) (From f) (To xk) (Amp a) (Width w) (Delay tau))) ) (Program set_dynamic_E (Args s t xk theta r L rk)  (Element (Time s−2dT) (Name xk) (Threshold −theta) (Refractory r)  (Last L))  (Meta (Name rk) (Window s t) (Element (Time t) (Name xk) (Threshold theta) (Refractory r) (Last L))) )

For D₀, follow the first row of separation table 4, reading the amplitudes from U, W, X, Y, Z to D₀ and the threshold for D₀. Observe that at time s-dT program set_dynamic_C initializes the amplitudes of the connections to A_(U,D0)=−2, A_(W,D0)=−2, A_(X,D0)=−2, A_(Y,D0)=2, A_(Z,D0)=2 as if R₀ doesn't fire. If R₀ does fire, then the Meta command in set_dynamic_C dynamically flips the sign of each of these amplitudes: at time t, the amplitudes are flipped to A_(U,D0)=2, A_(W,D0)=2, A_(X,D0)=2, A_(Y,D0)=−2, A_(Z,D0)=−2.

Similarly, the meta command in set_dynamic_E initializes the threshold of D₀ to θ_(D0)=−5 as if R₀ doesn't not fire. If R₀ does fire the meta command flips the sign of the threshold of D₀; for the D₀ case, the meta command sets θ_(D0)=5.

(set_dynamic_E s t D0 5 1 s−2 R0) (set_dynamic_C s t U D0 2 1 1 R0) (set_dynamic_C s t W D0 2 1 1 R0) (set_dynamic_C s t X D0 2 1 1 R0) (set_dynamic_C s t Y D0 −2 1 1 R0) (set_dynamic_C s t Z D0 −2 1 1 R0)

Similarly, for elements D₁, . . . , D₁₃, the commands set_dynamic_E and set_dynamic_C dynamically set the element parameters and the connections from U, X, W, Y, Z to D₁, . . . D₁₃ based on the rest of the quantum random firing pattern R₁, . . . , R₁₃.

(set_dynamic_E s t D1 3 1 s−2 R1) (set_dynamic_C s t U D1 2 1 1 R1) (set_dynamic_C s t W D1 2 1 1 R1) (set_dynamic_C s t X D1 −2 1 1 R1) (set_dynamic_C s t Y D1 −2 1 1 R1) (set_dynamic_C s t Z D1 −2 1 1 R1) (set_dynamic_E s t D2 5 1 s−2 R2) (set_dynamic_C s t U D2 2 1 1 R2) (set_dynamic_C s t W D2 2 1 1 R2) (set_dynamic_C s t X D2 −2 1 1 R2) (set_dynamic_C s t Y D2 −2 1 1 R2) (set_dynamic_C s t Z D2 2 1 1 R2) (set_dynamic_E s t D3 5 1 s−2 R3) (set_dynamic_C s t U D3 2 1 1 R3) (set_dynamic_C s t W D3 2 1 1 R3) (set_dynamic_C s t X D3 −2 1 1 R3) (set_dynamic_C s t Y D3 2 1 1 R3) (set_dynamic_C s t Z D3 −2 1 1 R3) (set_dynamic_E s t D4 3 1 s−2 R4) (set_dynamic_C s t U D4 2 1 1 R4) (set_dynamic_C s t W D4 −2 1 1 R4) (set_dynamic_C s t X D4 2 1 1 R4) (set_dynamic_C s t Y D4 −2 1 1 R4) (set_dynamic_C s t Z D4 −2 1 1 R4) (set_dynamic_E s t D5 5 1 s−2 R5) (set_dynamic_C s t U D5 2 1 1 R5) (set_dynamic_C s t W D5 −2 1 1 R5) (set_dynamic_C s t X D5 2 1 1 R5) (set_dynamic_C s t Y D5 −2 1 1 R5) (set_dynamic_C s t Z D5 2 1 1 R5) (set_dynamic_E s t D6 5 1 s−2 R6) (set_dynamic_C s t U D6 2 1 1 R6) (set_dynamic_C s t W D6 −2 1 1 R6) (set_dynamic_C s t X D6 2 1 1 R6) (set_dynamic_C s t Y D6 2 1 1 R6) (set_dynamic_C s t Z D6 −2 1 1 R6) (set_dynamic_E s t D7 1 1 s−2 R7) (set_dynamic_C s t U D7 2 1 1 R7) (set_dynamic_C s t W D7 −2 1 1 R7) (set_dynamic_C s t X D7 −2 1 1 R7) (set_dynamic_C s t Y D7 −2 1 1 R7) (set_dynamic_C s t Z D7 −2 1 1 R7) (set_dynamic_E s t D8 3 1 s−2 R8) (set_dynamic_C s t U D8 2 1 1 R8) (set_dynamic_C s t W D8 −2 1 1 R8) (set_dynamic_C s t X D8 −2 1 1 R8) (set_dynamic_C s t Y D8 2 1 1 R8) (set_dynamic_C s t Z D8 −2 1 1 R8) (set_dynamic_E s t D9 5 1 s−2 R9) (set_dynamic_C s t U D9 −2 1 1 R9) (set_dynamic_C s t W D9 2 1 1 R9) (set_dynamic_C s t X D9 2 1 1 R9) (set_dynamic_C s t Y D9 −2 1 1 R9) (set_dynamic_C s t Z D9 2 1 1 R9) (set_dynamic_E s t D10 5 1 s−2 R10) (set_dynamic_C s t U D10 −2 1 1 R10) (set_dynamic_C s t W D10 2 1 1 R10) (set_dynamic_C s t X D10 2 1 1 R10) (set_dynamic_C s t Y D10 2 1 1 R10) (set_dynamic_C s t Z D10 −2 1 1 R10) (set_dynamic_E s t D11 1 1 s−2 R11) (set_dynamic_C s t U D11 −2 1 1 R11) (set_dynamic_C s t W D11 2 1 1 R11) (set_dynamic_C s t X D11 −2 1 1 R11) (set_dynamic_C s t Y D11 −2 1 1 R11) (set_dynamic_C s t Z D11 −2 1 1 R11) (set_dynamic_E s t D12 3 1 s−2 R12) (set_dynamic_C s t U D12 −2 1 1 R12) (set_dynamic_C s t W D12 2 1 1 R12) (set_dynamic_C s t X D12 −2 1 1 R12) (set_dynamic_C s t Y D12 −2 1 1 R12) (set_dynamic_C s t Z D12 2 1 1 R12) (set_dynamic_E s t D13 5 1 s−2 R13) (set_dynamic_C s t U D13 −2 1 1 R13) (set_dynamic_C s t W D13 2 1 1 R13) (set_dynamic_C s t X D13 −2 1 1 R13) (set_dynamic_C s t Y D13 2 1 1 R13) (set_dynamic_C s t Z D13 2 1 1 R13)

Set up connections to active elements G₀, G₁, G₂, . . . G₁₄ which represent the number of elements in {R₀, R₁, R₂, . . . R₁₃} that are firing. If 0 are firing, then only G₀ is firing. Otherwise, if k>0 elements in {R₀, R₁, R₂, . . . R₁₃} are firing, then only G₁, G₂, . . . G_(k) are firing.

(Program firing_count (Args G a b theta) (Element (Time a−2dT) (Name G) (Threshold theta) (Refractory b−a) (Last 2a−b)) (Connection (Time a−dT) (From R0) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R1) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R2) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R3) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R4) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R5) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R6) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R7) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R8) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R9) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R10) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R11) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R12) (To G) (Amp 2) (Width b−a) (Delay 1)) (Connection (Time a−dT) (From R13) (To G) (Amp 2) (Width b−a) (Delay 1)) ) (firing_count G0 a b −1) (firing_count G1 a b 1) (firing_count G2 a b 3) (firing_count G3 a b 5) (firing_count G4 a b 7) (firing_count G5 a b 9) (firing_count G6 a b 11) (firing_count G7 a b 13) (firing_count G8 a b 15) (firing_count G9 a b 17) (firing_count G10 a b 19) (firing_count G11 a b 21) (firing_count G12 a b 23) (firing_count G13 a b 25) (firing_count G14 a b 27)

P₃ is the output of η3. Initialize element P₃'s threshold based on meta commands that use the information from elements G₀, G₁, . . . G₁₃. Observe that t+dT<t+2dT< . . . <t+10dT so the infinitesimal dT enables us to set the appropriate threshold for P₃ based on the number of firings. For example, if nine of the randomly chosen bits are high, then G₉ will fire, so the threshold of P₃ is set to 17. If five of the random bits are high, then the threshold of P₃ is set to 9.

(Program set_P_threshold (Args G P s t a b theta kdT)  (Meta (Name G) (Window s t) (Element (Time t+kdT) (Name P) (Threshold theta) (Refractory b−a) (Last t−b+a)) ) (set_P_threshold G0 P3 s t a b −1 dT) (set_P_threshold G1 P3 s t a b 1 2dT) (set_P_threshold G2 P3 s t a b 3 3dT) (set_P_threshold G3 P3 s t a b 5 4dT) (set_P_threshold G4 P3 s t a b 7 5dT) (set_P_threshold G5 P3 s t a b 9 6dT) (set_P_threshold G6 P3 s t a b 11 7dT) (set_P_threshold G7 P3 s t a b 13 8dT) (set_P_threshold G8 P3 s t a b 15 9dT) (set_P_threshold G9 P3 s t a b 17 10dT) (set_P_threshold G10 P3 s t a b 19 11dT) (set_P_threshold G11 P3 s t a b 21 12dT) (set_P_threshold G12 P3 s t a b 23 13dT) (set_P_threshold G13 P3 s t a b 25 14dT) (set_P_threshold G14 P3 s t a b 27 15dT)

Set up dynamical connections from D₀, D₁, . . . D₁₃ to P3 based on the random bits stored by R₀, R₁, . . . R₁₃. These connections are based on meta commands that use the firing pattern from elements R₀, R₁, . . . R₁₃.

(Program set_from_Xk_to_Pj (Args s t Xk Pj amp w tau Rk) (Connection (Time s−dT) (From Xk) (To Pj) (Amp −amp) (Width w) (Delay tau)) (Meta (Name Rk) (Window s t) (Connection (Time t) (From Xk) (To Pj) (Amp amp) (Width w) (Delay tau)) ) (set_from_Xk_to_Pj s t D0 P3 2 b−a 1 R0) (set_from_Xk_to_Pj s t D1 P3 2 b−a 1 R1) (set_from_Xk_to_Pj s t D2 P3 2 b−a 1 R2) (set_from_Xk_to_Pj s t D3 P3 2 b−a 1 R3) (set_from_Xk_to_Pj s t D4 P3 2 b−a 1 R4) (set_from_Xk_to_Pj s t D5 P3 2 b−a 1 R5) (set_from_Xk_to_Pj s t D6 P3 2 b−a 1 R6) (set_from_Xk_to_Pj s t D7 P3 2 b−a 1 R7) (set_from_Xk_to_Pj s t D8 P3 2 b−a 1 R8) (set_from_Xk_to_Pj s t D9 P3 2 b−a 1 R9) (set_from_Xk_to_Pj s t D10 P3 2 b−a 1 R10) (set_from_Xk_to_Pj s t D11 P3 2 b−a 1 R11) (set_from_Xk_to_Pj s t D12 P3 2 b−a 1 R12) (set_from_Xk_to_Pj s t D13 P3 2 b−a 1 R13)

Similar procedures use random firing patterns on active elements {A₀, A₁, . . . A₁₄}, {B₀, B₁, . . . B₁₅}, {C₀, C₁, . . . C₁₄}, {E₀, E₁, . . . E₁₂}, and {F₀, F₁, . . . F₁₃} to compute η₀, η₁, η₂, η₄ and η₅, respectively. The outputs of η₀, ƒ₁, η₂, η₄ and η₅ are represented by active elements P₀, P₁, P₂, P₄ and P₅, respectively. The level set rules for η₀, η₁, η₂, η₄ and η₅ are shown, respectively in tables 5, 6, 7, 8 and 9.

Since the firing activity of element P_(k) represents a single bit that helps determine the next state or next tape symbol during a UTM computational step, its firing activity and parameters can be assumed to remain perfectly secret. Alternatively, if an eavesdropper is able to listen to the firing activity of P₀, P₁, P₂, P₃, P₄ and P₅, which collectively represent the computation of η(UXW, YZ), then this leaking of information could be used to reconstruct some or all of the UTM tape contents.

This weakness can be rectified as follows. For each UTM computational step, the active element machine uses six additional quantum random bits b₀, b₁, b₂, b₃, b₄ and b₅. For element P₃, if random bit b₃=1, then the dynamical connections from D₀, D₁, . . . D₁₃ to P₃ are chosen as described above. However, if random bit b₃=0, then the amplitudes of the connections from D₀, D₁, . . . D₁₃ to P₃ and the threshold of P₃ are multiplied by −1. This causes P₃ to fire when η(UXW, YZ)=0 and P₃ doesn't fire when η(UXW, YZ)=1.

This cloaking of P₃'s firing activity can be coordinated with a meta command based on the value of b₃ so that P₃'s firing can be appropriately interpreted to dynamically change the active elements and connections that update the tape contents and state after each UTM computational step. This cloaking procedure can also be used for element P₀ and random bit b₀, P₁ and random bit b₁, P₂ and random bit b₂, P₄ and random bit b₄, and P₅ and random bit b₅.

Besides representing and computing the program η with quantum random firing patterns, there are other important functions computed by active elements executing the UTM. Assume that these connections and the active element firing activity are kept perfectly secret as they represent the state and the tape contents of the UTM tape contents. These functions are described below.

Three active elements (q 0), (q 1) and (q 2) store the current state of the UTM.

There are a collection of elements to represent the tape head location k where k is an integer.

A marker active element

locates the leftmost tape square and a separate marker active element

locates the rightmost tape square. Any tape symbols outside these markers are assumed to be blank i.e. 0. If the tape head moves beyond the leftmost tape square, then

's connection is removed and updated one tape square to the left and the machine is reading a 0. If the tape head moves beyond the rightmost tape square, then

's connection is removed and updated one tape square to the right and the machine is reading a 0.

There are a collection of elements that represent the tape contents of the UTM. For each tape square k inside the marker elements, there are two elements named (S k) and (T k) whose firing pattern determines the alphabet symbol at tape square k. For example, if elements (S 5) and (T 5) are not firing, then tape square 5 contains alphabet symbol 0. If element (S −7) is firing and element (T −7) is not firing, then tape square −7 contains alphabet symbol 1. If element (S 13) is firing and element (T 13) is firing, then tape square 13 contains alphabet symbol A.

Representing alphabet symbol 0 with two active elements that are not firing is convenient because if the tape head moves beyond the initial tape contents of the UTM, then the Meta command can add two elements that are not firing to represent the contents of the new square.

The copy program can be used to construct important functionality in the UTM. The following program enables a new alphabet symbol to be copied to the tape.

(Program copy_symbol (Args s t b0 a0 b1 a1) (copy s t b0 a0) (copy s t b1 a1) ) The following program enables a new state to be copied.

(Program copy_state (Args s t b0 a0 b1 a1 b2 a2) (copy s t b0 a0) (copy s t b1 a1) (copy s t b2 a2) )

The sequence of steps by which the UTM is executed with an AEM are described.

1. Tape contents are initialized and the marker elements L and R are initialized.

2. The tape head is initialized to tape square k=0 and the current machine state is initialized to q₂. In other words, (q 0) is not firing (q 1) is firing and (q 2) is not firing.

3. (S k) and (T k) are copied to a_(u), and the current state (q 0), (q 1), (q 2) is copied to q_(in).

4. η(q_(1n), a_(in))=(q_(out), a_(out), m) is computed where q_(out) represents the new state, a_(out) represents the new tape symbol and m represents the tape head move.

5. If

_(out)=h, then the UTM halts. The AEM reaches a static firing pattern that stores the current tape contents indefinitely and keeps the tape head fixed at tape square k where the UTM halted.

6. Otherwise, the firing pattern of the three elements representing q_(out) are copied to (q 0), (q 1), (q 2). a_(out) is copied to the current tape square represented by (S k), (T k).

7. If m=L, then first determine if the tape head has moved to the left of the tape square marked by

. If so, then have

remove its current marker and mark tape square k−1. In either case, go back to step 3 where (S k−1) and (T k−1) are copied to a_(in).

8. If m=R, then first determine if the tape head has moved to the right of the tape square marked by

. If so, then have

remove its current marker and mark tape square k+1. In either case, go back to step 3 where (S k+1) and (T k+1) are copied to a_(in).

Calude and Svozil showed that quantum randomness is Turing incomputable. Since the firing pattern of the active elements {A₀, A₁, . . . A₁₄} computing η₀; the firing pattern of elements {B₀, B₁, . . . B₁₅} computing η₁; the firing pattern of elements {C₀, C₁, . . . C₁₄} computing η₂; the firing pattern of elements {D₀, D₁, . . . D₁₃} computing η₃; the firing pattern of elements {E₀, E₁, . . . E₁₂ computing η₄; and the firing pattern of elements {F₀, F₁, . . . F₁₃} computing are all generated from quantum randomness, these firing patterns are Turing incomputable. As a consequence, there does not exist a Turing machine that can map these firing patterns back to the sequence of instructions executed by the universal Turing machine. In summary, these methods demonstrate a new class of computing machines where the purpose of the program's execution is incomprehensible (Turing incomputable) to malware.

Affine Fixed Points

Remark 1.1 Linear Maps as Matrices

The linear maps L and M are represented with matrices in the standard basis.

${L\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}\mspace{20mu} {M\begin{pmatrix} x \\ y \end{pmatrix}}} = {\begin{pmatrix} h & i \\ j & k \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}}$

Each affine transformation is of the form

${f\begin{pmatrix} x \\ y \end{pmatrix}} = {{L\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

where L or

$\quad\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

is called the linear part and

$\quad\begin{pmatrix} m \\ n \end{pmatrix}$

is called the translational part. The unit square with lower left corner at coordinate (j, k) is denoted as U_((j, k))={(x,y):j≦x≦j+1 & k≦y≦k+1} or it is sometimes represented with the corners of the square [(j, k), (j+1, k), (j+1, k+1), (j, k+1)].

Remark 1.3 Computation of the Composition of the Affine Maps

${g \circ {f\begin{pmatrix} x \\ y \end{pmatrix}}} = {{\begin{pmatrix} {{ah} + {ci}} & {{bh} + {di}} \\ {{aj} + {ck}} & {{bj} + {dk}} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} {{mh} + {ni} + p} \\ {{mj} + {nk} + q} \end{pmatrix}}$

Definition 1.4 Rectangular Area Preserving Affine Maps

An affine transformation r is called an rectangular area preserving affine transformation if it is of the form

${{r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{A} & 0 \\ 0 & A \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + \begin{pmatrix} m \\ n \end{pmatrix}}}\;$

for some A>0. In the next section, A is a number of the form B^(k), where k is an integer and B is an integer greater than 1, and m, n are rational numbers.

Remark 1.5 Rectangular Area Preserving Affine Maps are Closed.

Rectangular area preserving affine maps are closed under function composition. Consider

${r_{1}\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{A} & 0 \\ 0 & A \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + {\begin{pmatrix} m \\ n \end{pmatrix}\mspace{20mu} {and}}}$ ${r_{2}\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{B} & 0 \\ 0 & B \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + {\begin{pmatrix} p \\ q \end{pmatrix}\;.\begin{matrix} {{{{Then}\mspace{14mu} {r_{1} \circ {r_{2}\begin{pmatrix} x \\ y \end{pmatrix}}}} = {{\begin{pmatrix} \frac{1}{AB} & 0 \\ 0 & {AB} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + \begin{pmatrix} {\frac{m}{B} + p} \\ {{nB} + q} \end{pmatrix}}}\;} \\ {= {{\begin{pmatrix} \frac{1}{AB} & 0 \\ 0 & {AB} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + \begin{pmatrix} {\frac{p}{A} + m} \\ {{qA} + n} \end{pmatrix}}} \end{matrix}}}$

Remark 1.6

If bc+(1−a)(d−1)≠0 then

${f\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

has a fixed point at

$\left( {\frac{{\left( {d - 1} \right)m} - {bn}}{{bc} + {\left( {1 - a} \right)\left( {d - 1} \right)}},\frac{{\left( {a - 1} \right)n} - {cm}}{{bc} + {\left( {1 - a} \right)\left( {d - 1} \right)}}} \right).$

remark 1.7

When b=c=0 and (1−a)(d−1)≠0, then

${f\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

has a fixed point at

$\left( {\frac{m}{1 - a},\frac{n}{1 - d}} \right).$

Definition 1.8 Vertical Covers

Consider the rectangular area preserving affine map

${{r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{A} & 0 \\ 0 & A \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + \begin{pmatrix} m \\ n \end{pmatrix}}}\;$

where A>1 and with square domain W=[(u, v), (u+s, v), (u+s, v+s), (u, v+s)] with sides of length s. r (W) vertically covers W if and only if all four conditions A, B, C and D hold.

1.) The y-coordinate of r(u, v)<v.

2.) The y-coordinate of r(u+s, v+s)≧v+s.

C. The x-coordinate of r(u, v)≧u.

D. The x-coordinate of r(u+s, v+s)≦u+s.

See FIG. 14.

Remark 1.9 Vertical Cover Fixed Point

When a=1/A and d=A such that A>1, then

${{r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{A} & 0 \\ 0 & A \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}}\; + \begin{pmatrix} m \\ n \end{pmatrix}}}\;$

has a fixed point at

$\left( {\frac{m\; A}{A - 1},\frac{n}{1 - A}} \right).$

On square W=[(0,0), (s,0), (s,s), (0,s)], the image of affine map r on W is

${r(W)} = \left\lbrack {\left( {m,n} \right),\left( {{m + \frac{s}{A}},n} \right),\left( {{m + \frac{s}{A}},{n + {sA}}} \right),\left( {m,{n + {sA}}} \right)} \right\rbrack$

which is a rectangle with width

$\frac{s}{A}$

and height sA.

Definition 1.10 Horizontal Covers

Consider the rectangular area preserving affine map

${r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} A & 0 \\ 0 & \frac{1}{A} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

where A>1 and with square domain W=[(u, v), (u+s, v), (u+s, v+s), (u, v+s)] with sides of length s. r(W) horizontally covers W if and only if all four conditions hold:

1.) The x-coordinate of r(u, v)≦u.

2.) The x-coordinate of r(u+s, v+s)≧u+s.

C. The y-coordinate of r(u, v)≧v.

D. The y-coordinate of r(u+s, v+s)≦v+s.

See FIG. 15.

Remark 1.11 Horizontal Cover Fixed Point

When a=A and

$d = \frac{1}{A}$

such that A>1, then

${r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} A & 0 \\ 0 & \frac{1}{A} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

has a fixed point

$\left( {\frac{m}{1 - A},\frac{nA}{A - 1}} \right).$

On square W=[(0, 0), (s, 0), (s, s), (0, s)]

${r(W)} = \left\lbrack {{\left( {m,n} \right)\left( {{m + {sA}},n} \right)},\left( {{m + {sA}},{n + \frac{s}{A}}} \right),\left( {m,{n + \frac{s}{A}}} \right)} \right\rbrack$

is a rectangle with width sA and height

$\frac{s}{A}.$

Example 1.12

${r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} {1/2} & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} \frac{4}{10} \\ \frac{1}{10} \end{pmatrix}}$

with standard unit square U_((0, 0)). The fixed point

$\left( {\frac{8}{10},\frac{- 1}{10}} \right)$

is not in U_((0,0)). The geometry is shown in FIG. 16.

Theorem 1.13 Vertical Covering Fixed Point Theorem

Consider the rectangular, area-preserving affine map

${r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{A} & 0 \\ 0 & A \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

for some A>1 and square W=[(u, v), (u+s, v), (u+s, v+s) (u, v+s)]. Then r(W) vertically covers W if and only if fixed point

$\left( {\frac{mA}{A - 1},\frac{n}{1 - A}} \right)$

lies in W.

PROOF. Define

${r^{\prime}\begin{pmatrix} x \\ y \end{pmatrix}} = {r\begin{pmatrix} x \\ y \end{pmatrix}}$

to translate W so that W is a square of sides with length s with lower left corner at the origin. Thus, WLOG it can be assumed that the square domain W=[(0, 0), (s, 0), (s, s), (0, s)]. Observe that r(0, 0)=(m, n) and

${r\left( {s,s} \right)} = {\left( {{m + \frac{s}{A}},{n + {sA}}} \right).}$

See FIG. 17. Hence,

1.) The y-coordinate of r(0, 0)=n 2.) The y-coordinate of r(s, s)=n+sA 3.) The x-coordinate of r(0, 0)=m 4.) The x-coordinate of

${r\left( {s,s} \right)} = {m + {\frac{s}{A}\left( \Leftrightarrow \right)\left( {\frac{mA}{A - 1},\frac{n}{1 - A}} \right)}}$

lies in W if and only if

$0 \leq \frac{n}{1 - A} \leq {s\mspace{14mu} {AND}\mspace{14mu} 0} \leq \frac{mA}{A - 1} \leq s$

if and only if [n≦0 AND n≧(1−A)s] AND [0≦mA≦(A−1)s] if and only if

$\left\lbrack {n \leq {{0\mspace{14mu} {AND}\mspace{14mu} n} + {sA}} \geq s} \right\rbrack \mspace{14mu} {{AND}\mspace{14mu}\left\lbrack {0 \leq {{m\mspace{14mu} {AND}\mspace{14mu} m} + \frac{s}{A}} \leq s} \right\rbrack}$

if and only if [The y-coordinate of r(0, 0)≦0] AND [the y-coordinate of r(s, s)≦s] AND [the x-coordinate of r(0, 0)≧0] AND [the x-coordinate of r(s, s)≦s]

Theorem 1.14 Horizontal Covering Fixed Point Theorem

Consider the rectangular, area-preserving affine map

${r\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} A & 0 \\ 0 & \frac{1}{A} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

for some A>1 and square W=[(u, v), (u+s, v), (u+s, v+s) (u, v+s)].

Then r(W) horizontally covers W if and only if fixed point

$\left( {\frac{m}{1 - A},\frac{nA}{A - 1}} \right)$

lies in W.

PROOF. Define

${r^{\prime}\begin{pmatrix} x \\ y \end{pmatrix}} = {{r\begin{pmatrix} x \\ y \end{pmatrix}} - \begin{pmatrix} u \\ v \end{pmatrix}}$

to translate W so that W is the square with lower left corner at the origin. Thus, WLOG it can be assumed that W=[(0, 0), (s, 0), (s, s), (0, s)].

${r\left( {s,s} \right)} = {\left( {{m + {sA}},{n + \frac{s}{A}}} \right).}$

Observe r(0, 0)=(m, n) and See FIG. 18.

Hence,

1.) The x-coordinate of r(0, 0)=m

2.) The x-coordinate of r(s, s)=m+sA

3.) The y-coordinate of r(0, 0)=n

4.) The y-coordinate of

${r\left( {s,s} \right)} = {n + {\frac{s}{A}\left( \Leftrightarrow \right)\left( {\frac{m}{1 - A},\frac{nA}{A - 1}} \right)}}$

lies in W if and only if

$0 \leq \frac{m}{1 - A} \leq {s\mspace{14mu} {and}\mspace{14mu} 0} \leq \frac{nA}{A - 1} \leq s$

if and only if [m≦0 AND m≧A)s] and [0≦nA≦(A−1)s] if and only if

$\left\lbrack {m \leq {{0\mspace{14mu} {AND}\mspace{14mu} m} + {sA}} \geq s} \right\rbrack \mspace{14mu} {{and}\mspace{14mu}\left\lbrack {0 \leq {{n\mspace{14mu} {AND}\mspace{14mu} n} + \frac{s}{A}} \leq s} \right\rbrack}$

if and only if [The x-coordinate of r(0, 0)≦0] AND [the x-coordinate of r(s, s)≧s]

AND [the y-coordinate of r(0, 0)≧0] AND [the y-coordinate of r(s, s)≦s].

Turing Machine and Affine Map Correspondence

Definition 2.1 Turing Machine

A Turing machine is a triple (Q, A, η) where

-   -   There is a unique state h, called the halt state.     -   Q is a finite set of states that does not contain the halt         state h. The states are represented as Q={q₁, q₂, . . . q_(K)}         or as the natural numbers Q={2, . . . , K} and the halt state         as 1. Before machine execution begins, there is a starting state         s and s is an element of Q.     -   L and R are special symbols that instruct the machine to advance         to the left square or to the right square on the tape T.     -   A is a finite set of alphabet symbols that are read from and         written to the tape. The alphabet symbols are denoted as A={a₁,         a₂, . . . , a_(J)} or as the natural numbers A={1, 2, . . . ,         J}. A does not contain the symbols L, R.     -   η is a function where η: Q×A→(Q∪{h})×A×{L, R}.

The η function acts as the program for the Turing machine in the following manner. For each q in Q and a in A, the expression η(q, a)=(r, β, x) describes how machine (Q, A, η) executes one computational step. When in state q and scanning alphabet symbol a on the tape:

-   -   1.) Machine (Q, A, η) changes to state r.     -   2.) Machine (Q, A, η) rewrites alphabet symbol a as symbol β on         the tape.     -   3.) If x=L, then machine (Q, A, η) moves its tape head one         square to the left on the tape and is subsequently scanning the         symbol in this square.     -   4.) If x=R, then machine (Q, A, η) moves its tape head one         square to the right on the tape and is subsequently scanning the         symbol in this square.

5.) If r=h, machine (Q, A, η) enters the halting state h, and the machine stops (halts).

Definition 2.2 Turing Machine Tape

The Turing machine tape T is represented as a function T: Z→A where Z denotes the integers. The tape T is M-bounded if there exists a bound M>0 such that for T(k)=T(j) whenever |j|, |k|>M. (In some cases, the blank symbol # is used and T(k)=# when |k|>M) The symbol on the kth square of the tape is denoted as T_(k). Here we do not assume the tape is M-bounded unless it is explicitly stated for particular cases.

Definition 2.3 Turing Machine Configuration with Tape Head Location

Let (Q, A, η) be a Turing machine with tape T. A configuration is an element of the set

=(Q∪{h})×Z×{T: T is tape with range A}. The standard definition of a Turing machine assumes the initial tape is M-bounded and the tape contains only blank symbols, denoted as #, outside the bound.

If (q, k, T) is a configuration in

, then k is called the tape head location. The tape head location is M-bounded if there exists a natural number M>0 such that the tape head location k satisfies |k|≦M. A configuration whose first coordinate equals h is called a halted configuration. The set of non-halting configurations is

={(q, k, T)ε

: q≠h}

The purpose of the definition of a configuration is that the first coordinate stores the current state of the Turing machine, the third coordinate stores the contents of the tape, and the second coordinate stores the location of the tape head on the tape. Before presenting some examples of configurations, it is noted that there are different methods to describe the tape contents. One method is

${T(k)} = {\begin{Bmatrix} \alpha_{k} & {{{if}\mspace{14mu} l} \leq k \leq n} \\ \# & {otherwise} \end{Bmatrix}.}$

This is a max{|l|, |n|}-bounded tape. Another convenient representation is to list the tape contents and underline the symbol to indicate the location of the tape head. ( . . . ##αβ## . . . ).

A diagram can also represent the tape, tape head location, and the configuration (q, k, T). See FIG. 19.

Example 2.4 Turing Machine Configuration

Consider configuration (p, 2, . . . ##αβ## . . . ). The first coordinate indicates that the Turing machine is in state p. The second coordinate indicates that its tape head is currently scanning tape square 2, denoted as T₂ or T(2). The third coordinate indicates that tape square 1 contains symbol α, tape square 2 contains symbol β, and all other tape squares contain the # symbol.

Example 2.5 Halt Configuration Represented as Natural Numbers

A second example of a configuration is (1, 6, . . . 1111233111 . . . ). This configuration is a halted configuration. The first coordinate indicates that the machine is in halt state 1. The second coordinate indicates that the tape head is scanning tape square 6. The underlined 2 in the third coordinate indicates that the tape head is currently scanning a 2. In other words, T(6)=2, T(7)=3, T(8)=3, and T(k)=1 when k<6 OR k>8.

Definition 2.6 Turing Machine Computational Step

Consider machine (Q, A, η) with configuration (q, k, T) such that T(k)=α. After the execution of one computational step, the new configuration is one of the three cases such that for all three cases S(k)=β and S(j)=T(j) whenever j≠k:

Case I. (r, k−1, 5) if η(q, α)=(r,β, L).

Case II. (r, k+1, S) if η(q, α)=(r, β, R).

Case III. (h, k, 7). In this case, the machine execution stops (halts).

If the machine is currently in configuration (q₀, k₀, T₀) and over the next n steps the sequence of machine configurations (points) is (q₀, k₀, T₀), (q_(n), k_(n), T_(n)), . . . , (q_(n), k_(n), T_(n)) then this execution sequence is sometimes called the next n+1 computational steps.

If Turing machine (Q, A, η) with initial configuration (s, k, T) reaches the halt state h after a finite number of execution steps, then the machine execution halts. Otherwise, it is said that the machine execution is immortal on initial configuration (s, k, T).

The program symbol η induces a map η:

→

where η(q, k, T)=(r, k−1, S) when η(q, α)=(r, β, L) and η(q, k, T)=(r, k+1, S) when η(q, α)=(r, β, R).

Definition 2.7 Turing Program Length

The program length is the number of elements in the domain of η. The program length is denoted as |η|. Observe that |η|=|Q×A|=|Q∥A|. Note that in [7] and [32], they omit quintuples (q, a, r, b, x) when r is the halting state. In our representation, η(q, a)=(1, b, x) or η(q, a)=(h, b, x).

Definition 2.8 Tape Head glb

, lub

, Window of Execution [

,

]

Suppose a Turing machine begins or continues its execution with tape head at tape square k. During the next N computational steps, the greatest lower bound

of the tape head is the left most (smallest integer) tape square that the tape head visits during these N computational steps; and the least upper bound

of the tape head is the right most (largest integer) tape square that the tape head visits during these N computational steps. The window of execution denoted as [

,

] or [

,

+1, . . . ,

−1,

] is the sequence of integers representing the tape squares that the tape head visited during these N computational steps. The length of the window of execution is

−

+1 which is also the number of distinct tape squares visited by the tape head during these N steps. To express the window of execution for the next n computational steps, the lower and upper bounds are expressed as a function of n: [

(n),

(n)].

Example 2.9 Q={q, r, s, t, u, v, w, x}. A={1, 2}. Halting state=h

η(q, 1) = (r, 1, R). η(q, 2) = (h, 2, R). η(r, 1) = (h, 1, R). η(r, 2) = (s, 2, R). η(s, 1) = (t, 1, R). η(s, 2) = (h, 2, R). η(t, 1) = (h, 1, R). η(t, 2) = (u, 2, R). η(u, 1) = (h, 1, R). η(u, 2) = (v, 1, R). η(v, 1) = (h, 1, R). η(v, 2) = (w, 2, R). η(w, 1) = (h, 1, R). η(w, 2) = (x, 1, L). η(x, 1) = (h, 1, R). η(x, 2) = (q, 2, R). Left pattern = 12. Spanning Middle Pattern = 121 2212. Right pattern = 212.

The machine execution steps are shown in FIG. 20 with tape head initially at square 1. The tape head location is underlined. The tape head moves are {R⁶LR}^(n). The point p=[q, 12

1

212222] is an immortal periodic point with period 8 and hyperbolic degree 6.

Remark 2.10 If j≦k, then [

(j),

(j)]⊂[

(k),

(k)]

This follows immediately from the definition of the window of execution.

Since the tape squares may be renumbered without changing the results of the machine execution, for convenience it is often assumed that the machine starts execution at tape square 0. In example 2.9, during the next 8 computational steps—one cycle of the immortal periodic point—the window of execution is [0, 6]. The length of the window of execution is 7: Observe that if the tape squares are renumbered and the goal is to refer to two different windows of execution, for example [

(j),

(j)] and [

(k),

(k)], then both windows are renumbered with the same relative offset so that the two windows can be compared.

Definition 2.11 Value Function and Base

Suppose the alphabet A={a₁, a₂, . . . , a_(J)} and the states are Q={q₁, q₂, . . . q_(K)}. Define the symbol value function v: A∪Q∪{h}→N where N denotes the natural numbers. v(h)=0. v(a_(k))=k. v(q_(k))=k+|A|. v(q_(K))=|Q|+|A|. Choose the number base B=|Q|+|A|+1. Observe that 0≦v(x)<B and that each symbol chosen from A∪Q∪{h} has a unique value in base B.

Definition 2.12 Turing Machine Program Isomorphism

Two Turing machines M₁(Q₁, A₁, η₁) and M₂(Q₂, A₂, η₂) have a program isomorphism denoted as Ψ: M₁→M₂ if

-   -   A. There is a function φ: Q₁→Q₂ that is a bijection.     -   B. There is a function γ: A₁→A₂ that is a bijection.     -   C. There is a function Ψ: M₁→M₂ such that for each pair (q, a)         in Q₁×A₁ Ψ(η₁(q, a))=η₂(φ(q), γ(a))

Remark 2.13 If alphabet A={a}, then the halting behavior of the Turing machine is completely determined in ≦|Q|+1 execution steps.

PROOF. Suppose Q={q₁, q₂, . . . q_(K)}. Observe that the program length is |η|=|Q|. Also, after an execution step every tape symbol on the tape must be a. Consider the possible execution steps: η(q_(S(1)), a)→η(q_(S(2)), a)→η(q_(S(3)), a) . . . →η(q_(S(K+1)), a). If the program execution does not halt in these |Q|+1 steps, then S(i)=S(j) for some i≠j; and the tape contents are still all a's. Thus, the program will exhibit periodic behavior whereby it will execute η(q_(S(i)), a)→ . . . →η(q_(S(j)), a) indefinitely. If the program does not halt in |Q|+1 execution steps, then the program will never halt.

As a result of Remark 2.13, from now on, it is assumed that |A|≧2. Further, since at least one state is needed, then from here on, it is assumed that the base B≧3.

Definition 2.14 Turing Machine Configuration to x-y plane P correspondence. See FIG. 19. For a fixed machine (Q, A, η), each configuration (q, k, T) represents a unique point (x, y) in P. The coordinate functions x:

→P and y:

→P, where

is the set of configurations, are x(q, k, T)=T_(k) T_(k+1). T_(k+2) T_(k+3) T_(k+4) . . . where this decimal sequence in base B represents the real number

${{{Bv}\left( T_{k} \right)} + {v\left( T_{k + 1} \right)} + {\sum\limits_{j = 1}^{x}{{v\left( T_{k + j + 1} \right)}B^{- j}{y\left( {q,k,T} \right)}}}} = {{{qT}_{k - 1} \cdot T_{k - 2}}T_{k - 3}T_{k - 4}\mspace{14mu} \ldots}$

where this decimal sequence in base B represents a real number as

${{Bv}(q)} + {v\left( T_{k - 1} \right)} + {\sum\limits_{j = 1}^{\propto}{{v\left( T_{k - j - 1} \right)}B^{- j}}}$

Define function φ:

→P as φ(q, k, T) (x(q, k, T), y(q, k, T)). φ is the function that maps machine configurations into points into the plane P.

Definition 2.15 Equivalent Configurations

With respect to Turing machine (Q, A, η), the two configurations (q, k, T) and (q, j, V) are equivalent [i.e. (q, k, T)˜(q, j, V)] if T(m)=V(m+j−k) for every integer m. Observe that ˜ is an equivalence relation on

. Let

′ denote the set of equivalence classes [(q, k, T)] on

Also observe that φ maps every configuration in equivalence class [(q, k, T)] to the same ordered pair of real numbers in P. Recall that (D, X) is a metric space if the following three conditions hold.

1. D(a, b)≧0 for all a, b in X where equality holds if and only if a=b.

2. D(a, b)=D(b, a) for all a, b. (Symmetric)

3. D(a, b)=D(a, c)+D(c, b) for all a, b, c in X. (Triangle inequality.)

(ρ,

′) is a metric space where ρ is induced via φ by the Euclidean metric in P. Consider points p₁, p₂ in P with p=(x₁, y₁) and p₂=(x₂, y₂) where (d, P) is a metric space with Euclidean metric d(p₁, P₂)=√{square root over ((x₁−x₂)²+(y₁−y₂)²)}{square root over ((x₁−x₂)²+(y₁−y₂)²)}

Let u=[(q, k, S)], w=[(r, l, T)] be elements of

′. Define ρ:

′×

′→R as ρ(u, w)=d (φ(u),

$\left. {\phi (w)} \right) = \sqrt{\left\lbrack {{x\left( {q,k,S} \right)} - {x\left( {r,l,T} \right)}} \right\rbrack^{2} + \left\lbrack {y\left( {r,l,T} \right)} \right\rbrack^{2}}$

The symmetric property and the triangle inequality hold for ρ because d is a metric. In regard to property (i), ρ(u, w)≧0 because d is a metric. The additional condition that ρ(u, w)=0 if and only if u=w holds because d is a metric and because the equivalence relation—collapses non-equal equivalent configurations (i.e. two configurations in the same state but with different tape head locations and with all corresponding symbols on their respective tapes being equal) into the same point in

′.

The unit square U_((└x┘, └y┘)) has a lower left corner with coordinates (└x┘, └y┘) where └x┘=Bv(T_(k))+v(T_(k+1)) and └y┘=Bv(q)+v(T_(k−1)). See FIG. 21.

Definition 2.16 Left Affine Function

This is for case I. where η(q, T_(k))=(r, β, L). See FIG. 22.

x

T _(k−1) β·T _(k+1) T _(k+2) T _(k+3) . . .

B ⁻¹ x=T _(k) ·T _(k+1) T _(k+2) T _(k+3) T _(k+4)

Thus, m=T_(k−1)β−T_(k) where the subtraction of integers is in base B.

y

rT _(k−2) ·T _(k−3) T _(k−4) T _(k−5) . . .

By=qT _(k−1) T _(k−2) ·T _(k−3) T _(k−4) T _(k−5) . . .

Thus, n=rT_(k−2)−qT_(k−1)T_(k−2) where the subtraction of integers is in base B.

Define the left affine function F_((└x┘, └y┘)): U_((└x┘, └y┘))→P where

${{F_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{B} & 0 \\ 0 & B \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}},$

m=Bv(T_(k−1))+v(β)−v(Tk) and n=Bv(r)−B²v(q)−Bv(T_(k−1)).

LEMMA 2.17 Left Affine Function

Turing Machine Computational Step

Let (q, k, T) be a Turing machine configuration. Suppose η(q, T_(k))=(r, b, L) for some state r in Q∪{h} and some alphabet symbol b in A and where T_(k)=a. Consider the next Turing Machine computational step. The new configuration is (r, k−1, T^(b)) where T^(b)(j)=T(j) for every j≠k and T^(b)(k)=b. The commutative diagram φη(q, k, T)=F_((└x┘, └y┘)) φ(q, k, T) holds. In other words, F_((└x┘, └y┘)) [x(q, k, T), y(q, k, T)]=[x(r, k−1, T^(b)), y(r, k−1, T^(b))].

PROOF. x(r, k−1, T^(b))=T_(k−1)b·T_(k+1) T_(k+2) . . .

The x coordinate of

$\begin{matrix} {{F_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\left\lbrack {{x\left( {q,k,T} \right)},{y\left( {q,k,T} \right)}} \right\rbrack} = {{B^{- 1}{x\left( {q,k,T} \right)}} + {{Bv}\left( T_{k - 1} \right)} + {v(b)} - {v(a)}}} \\ {= {{B^{- 1}\left( {{{aT}_{k + 1} \cdot T_{k + 2}}T_{k + 3}\mspace{14mu} \ldots}\mspace{14mu} \right)} + {{Bv}\left( T_{k - 1} \right)} +}} \\ {{{v(b)} - {v(a)}}} \\ {= {{{a \cdot T_{k + 1}}T_{k + 2}T_{k + 3}\mspace{14mu} \ldots}\mspace{14mu} + {{Bv}\left( T_{k - 1} \right)} +}} \\ {{{v(b)} - {v(a)}}} \\ {= {T_{k - 1}{b \cdot T_{k + 1}}T_{k + 2}T_{k + 3}\mspace{14mu} \ldots}} \end{matrix}$      y(r, k − 1, T^(b)) = rT_(k − 2) ⋅ T_(k − 3)T_(k − 4)  …

The y coordinate of

$\begin{matrix} {{F_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\left\lbrack {{x\left( {q,k,T} \right)},{y\left( {q,k,T} \right)}} \right\rbrack} = {{{By}\left( {q,k,T} \right)} + {{Bv}(r)} - {B^{2}{v(q)}} - {{{Bv}\left( T_{k - 1} \right)}.}}} \\ {= {{B\left( {{{qT}_{k - 1} \cdot T_{k - 2}}T_{k - 3}\mspace{14mu} \ldots}\mspace{14mu} \right)} + {{Bv}(r)} - {B^{2}{v(q)}} -}} \\ {{{{Bv}\left( T_{k - 1} \right)}.}} \\ {= {{{qT}_{k - 1}{T_{k - 2} \cdot T_{k - 3}}\mspace{14mu} \ldots} + {{Bv}(r)} - {B^{2}{v(q)}} -}} \\ {{{{Bv}\left( T_{k - 1} \right)}.}} \\ {= {{{rT}_{k - 2} \cdot T_{k - 3}}T_{k - 4}\mspace{14mu} \ldots}} \\ {= {{y\left( {r,{k - 1},T^{b}} \right)}.}} \end{matrix}$

REMARK 2.18 Minimum vertical translation for left affine function As in 2.16, n is the vertical translation. |Bv(r)−Bv(T_(k−1))|=B|v(r)−v(T_(k−1))|≦B(B−1). Since q is a state, v(q)≧(|A|+1). This implies |−−B²v(q)|≧(|A|+1)B² This implies that |n|≧(|A|+1)B²−B(B−1)≧|A|B²+B.

Thus, |n|≧|A|B²+B.

Definition 2.19 Right Affine Function

This is for case II. where η(q, T_(k))=(r, β, R). See FIG. 23.

x

T _(k+1) T _(k+2) ·T _(k+3) T _(k+4) . . .

Bx=T _(k) T _(k+1) T _(k+2) ·T _(k+3) T _(k+4) . . .

Thus, m=T_(k+1)T_(k+2)−T_(k)T_(k+1)T_(k+2) where the subtraction of integers is in base B.

y

rβ·T _(k−1) T _(k−2) T _(k−3) . . .

B ⁻¹ y=q·T _(k−1) T _(k−2) T _(k−3) . . .

Thus, n=rβ−q where the subtraction of integers is in base B.

Define the right affine function G_((└x┘, └y┘)): U_((└x┘, └y┘))→P such that

${g_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} B & 0 \\ 0 & \frac{1}{B} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m \\ n \end{pmatrix}}$

where m=−B²v(T_(k)) and n=Bv(r)+v(β)−v(q).

LEMMA 2.20 Right Affine Function

Turing Machine Computational Step

Let (q, k, T) be a Turing machine configuration. Suppose η(q, T_(k))=(r, b, R) for some state r in Q∪{h} and some alphabet symbol b in A and where T_(k)=a. Consider the next Turing Machine computational step. The new configuration is (r, k+1, T^(b)) where T^(b)(j)=T(j) for every j≠k and T^(b)(k)=b. The commutative diagram  η(q, k, T)=G_((└x┘, └y┘))φ(q, k, T) holds.

In other words, G_((└x┘, └y┘)) [x(q, k, T), y(q, k, T)]=[x(r, k+1, T^(b)), y(r, k+1, T^(b))].

PROOF. From η(q, T_(k))=(r, b, R), it follows that x(r, k+1, T^(b))=T_(k+1) T_(k+2)·T_(k+3) T_(k+4) . . .

The x coordinate of

$\begin{matrix} {{G_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\left\lbrack {{x\left( {q,k,T} \right)},{y\left( {q,k,T} \right)}} \right\rbrack} = {{{Bx}\left( {q,k,T} \right)} - {B^{2}{v(a)}}}} \\ {= {{B\left( {{{aT}_{k + 1} \cdot T_{k + 2}}T_{k + 3}T_{k + 4}\mspace{14mu} \ldots}\mspace{14mu} \right)} - {B^{2}{v(a)}}}} \\ {= {{{aT}_{k + 1}{T_{k + 2} \cdot T_{k + 3}}T_{k + 4}\mspace{14mu} \ldots}\mspace{14mu} - {B^{2}{v(a)}}}} \\ {= {T_{k + 1}{T_{k + 2} \cdot T_{k + 3}}T_{k + 4}}} \\ {= {x\left( {r,{k + 1},T^{b}} \right)}} \end{matrix}$

From η(q, T_(k))=(r, b, R), it follows that y(r, k+1, T^(b))=rb·T_(k−1)T_(k−2)T_(k−3) . . .

The y coordinate of

$\begin{matrix} {{G_{({{\lfloor x\rfloor},{\lfloor y\rfloor}})}\left\lbrack {{x\left( {q,k,T} \right)},{y\left( {q,k,T} \right)}} \right\rbrack} = {{B^{- 1}{y\left( {q,k,T} \right)}} + {{Bv}(r)} + {v(b)} - {v(q)}}} \\ {= {{B^{- 1}\left( {{{qT}_{k - 1} \cdot T_{k - 2}}T_{k - 3}\mspace{14mu} \ldots}\mspace{14mu} \right)} + {{Bv}(r)} +}} \\ {{{v(b)} - {v(q)}}} \\ {= {{{q \cdot T_{k - 1}}T_{k - 2}T_{k - 3}\mspace{14mu} \ldots} + {{Bv}(r)} + {v(b)} - {v(q)}}} \\ {= {{{rb} \cdot T_{k - 1}}T_{k - 2}T_{k - 3}\mspace{14mu} \ldots}} \\ {= {{y\left( {r,{k + 1},T^{b}} \right)}.}} \end{matrix}$

REMARK 2.21 Minimum vertical translation for right affine function First,

v(β) − v(q) ≤ B − 1. $\begin{matrix} {{n} = {{{{Bv}(r)} + {v(\beta)} - {v(q)}}}} \\ {\geq {{{Bv}(r)} - \left( {B - 1} \right)}} \\ {\geq \left( {{A} + {1B} - \left( {B - 1} \right)} \right.} \end{matrix}$ because $\begin{matrix} {{v(r)} \geq {{A} + 1}} \\ {= {{{A}B} + 1.}} \end{matrix}$

Thus, |n|≧|A|B+1.

Definition 2.22 Function Index Sequence and Function Sequence

Let {ƒ₁, ƒ₂, . . . , ƒ_(I)} be a set of functions such that each function ƒ_(k): X→X. Then a function index sequence is a function S: N→{1, 2, . . . I} that indicates the order in which the functions {ƒ₁, ƒ₂, . . . , ƒ_(I)} are applied. If p is a point in X, then the orbit with respect to this function index sequence is [p, ƒ_(S(1))(p), ƒ_(S(2)),ƒ_(S(1))(p), . . . , ƒ_(S(m)ƒ) _(S(m−1)) . . . ƒ_(S(2))ƒ_(S(1))(p), . . . ]. Square brackets indicate that the order matters. Sometimes the first few functions will be listed in a function sequence when it is periodic. For example, [η₀, η₁, η₀, η₁, . . . ] is written when formally this function index sequence is S: N→{0, 1} where S(n)=n mod 2.

Example 2.23

${f\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} 4 \\ 0 \end{pmatrix}}$

on domain U_((0, 0)) and

${g\begin{pmatrix} x \\ y \end{pmatrix}} = {{\begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} {- 1} \\ 0 \end{pmatrix}}$

on U_((4, 0))

${g \circ {f\begin{pmatrix} x \\ y \end{pmatrix}}} = {{\begin{pmatrix} \frac{1}{16} & 0 \\ 0 & 16 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + {\begin{pmatrix} 0 \\ 0 \end{pmatrix}\mspace{14mu} {and}}}$ $\begin{matrix} {{f\begin{pmatrix} 0 \\ 0 \end{pmatrix}} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}} & {{f\begin{pmatrix} 1 \\ 0 \end{pmatrix}} = \begin{pmatrix} 4.25 \\ 0 \end{pmatrix}} & {{f\begin{pmatrix} 0 \\ 1 \end{pmatrix}} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}} \end{matrix}$

(0, 0) is a fixed point of gƒ. The orbit of point p chosen from the horizontal segment connected by the points (0, 0) and (1,0) with respect to the function sequence [ƒ, g, ƒ, g, . . . ] is a subset of U_((0, 0))∪U_((4, 0)). The point p is called an immortal point. The orbit of a point Q outside this segment exits (halts) U_((0,0))∪U_((4, 0)).

Definition 2.24 Halting and Immortal Orbits in the Plane.

Let P denote the two dimensional x,y plane. Suppose ƒ_(k): U_(k)→P is a function for each k such that whenever j≠k, then U_(j)∩U_(k)=Ø. For any point p in the plane P an orbit may be generated as follows. The 0^(th) iterate of the orbit is p. Given the kth iterate of the orbit is point q, if point q does not lie in any U_(k), then the orbit halts. Otherwise, q lies in at least one U_(j). Inductively, the k+1 iterate of q is defined as ƒ_(j)(q). If p has an orbit that never halts, this orbit is called an immortal orbit and p is called an immortal point. If p has an orbit that halts, this orbit is called a halting orbit and p is called a halting point.

Theorem 2.25 Turing Machine Execution

Affine Map Orbit Halting/Immortal Orbit Correspondence Theorem

Consider Turing machine (Q, A, η) with initial tape configuration (s, 0, T). W.L.O.G., it is assumed that the machine begins executing with the tape head at zero. Let ƒ₁, ƒ₂, . . . , ƒ_(I) denote the I affine functions with corresponding unit square domains W₁, W₂, W₃, . . . , W_(i) determined from 2.14, 2.15, 2.16 and 2.19. Let p=(x(s, 0, T), y(s, 0, T). Then from 2.14,

${x\left( {s,0,T} \right)} = {{{Bv}\left( T_{0} \right)} + {v\left( T_{1} \right)} + {\sum\limits_{j = 1}^{\propto}{{v\left( T_{j + 1} \right)}{B^{- j}.}}}}$

Also,

${y\left( {s,0,T} \right)} = {{{Bv}(s)} + {v\left( T_{- 1} \right)} + {\sum\limits_{j = 1}^{\propto}{{v\left( T_{{- j} - 1} \right)}{B^{- j}.}}}}$

There is a 1 to 1 correspondence between the mth point of the orbit

$\left\lbrack {p,{f_{S{(1)}}(p)},{f_{S{(2)}}{f_{S{(1)}}(p)}},\ldots \mspace{14mu},{f_{S{(m)}}f_{S{({m - 1})}}\mspace{14mu} \ldots \mspace{14mu} f_{S{(2)}}{f_{S{(1)}}(p)}},\ldots}\mspace{14mu} \right\rbrack \subseteq {\bigcup\limits_{k = 1}^{I}W_{k}}$

and the mth computational step of the Turing machine (Q, A, η) with initial configuration (s, 0, T). In particular, the Turing Machine halts on initial configuration (s, 0, T) if and only if p is a halting point with respect to affine functions ƒ_(k): W_(k)→P where 1≦k≦I. Dually, the Turing Machine is immortal on initial configuration (s, 0, T) if and only if p is an immortal point with respect to affine functions ƒ_(k): W_(k)→P where 1≦k≧I.

PROOF. From lemmas 2.17, 2.20, definition 2.14 and remark 2.15, every computational step of (Q, A, η) on current configuration (q, k, T′) corresponds to the application of one of the unique affine maps ƒ_(k), uniquely determined by remark 2.15 and definitions 2.16, 2.19 on the corresponding point p=[x(r, k, T′), y(r, k, T′)]. Thus by induction, the correspondence holds for all n if the initial configuration (s, 0, T) is an immortal configuration which implies that [x(s, 0, T), y(s, 0, T)] is an immortal point. Similarly, if the initial configuration (s, 0, T) is a halting configuration, then the machine (Q, A, η) on (s, 0, T) halts after N computational steps. For each step, the correspondence implies that the orbit of initial point p₀=[x(s, 0, T), y(s, 0, T)] exits

$\bigcup\limits_{k = 1}^{I}W_{k}$

on the Nth iteration of the orbit. Thus, p₀ is a halting point.

Corollary 2.26 Immortal Periodic Points, Induced by Configurations, Correspond to Equivalent Configurations that are Immortal Periodic.

PROOF. Suppose p=[x(q, k, T), y(q, k, T)] with respect to (Q, A, η) and p lies in

$\bigcup\limits_{k = 1}^{I}W_{k}$

such that ƒ_(S(N))ƒ_(S(N−1)) . . . ƒ_(S(1))(p)=p. Starting with configuration (q, k, T), after N execution steps of (Q, A, η), the resulting configuration (q, j, V) satisfies x(q, k, T)=x(q, j, V) and y(q, k, T)=y(q, j, V) because of ƒ_(S(N))ƒ_(S(N−1)) . . . η_(S(1))(p)=p and Theorem 2.25. This implies that (q, k, T) is translation equivalent to (q, j, V).

By induction this argument may be repeated indefinitely. Thus, (q, k, T) is an immortal configuration such that for every N computational steps of (Q, A, η), the kth resulting configuration (q, j_(k), V_(k)) is translation equivalent to (q, k, T).

Corollary 2.27 Immortal Periodic Points, Induced by Configurations, Correspond to Equivalent Configurations that are Immortal Periodic.

PROOF. Suppose p=[x(q, k, T), y(q, k, T)] with respect to (Q, A, η) and p lies in

$\bigcup\limits_{k = 1}^{I}W_{k}$

such that η_(S(N))ƒ_(S(N−1)) . . . ƒ_(S(1))(p)=p. Starting with configuration (q, k, T), after N execution steps of (Q, A, η), the resulting configuration (q, j, V) satisfies x(q, k, T)=x(q, j, V) and y(q, k, T)=y(q, j, V) because of ƒ_(S(N))ƒ_(S(N−1)) . . . ƒ_(S(1))(p)=p and Theorem 2.25. This implies that (q, k, T) is translation equivalent to (q, j, V).

By induction this argument may be repeated indefinitely. Thus, (q, k, T) is an immortal configuration such that for every N computational steps of (Q, A, η), the kth resulting configuration (q, j_(k), V_(k)) is translation equivalent to (q, k, T).

LEMMA 2.28 Two affine functions with adjacent unit squares as their respective domains are either both right affine or both left affine functions. (Adjacent unit squares have lower left x and y coordinates that differ at most by 1. It is assumed that |Q|≧2, since any Turing program with only one state has a trivial halting behavior that can be determined in |A| execution steps when the tape is bounded.)

PROOF. The unit square U_((└x┘, └y┘)) has a lower left corner with coordinates (└x┘, └y┘) where └x┘=Bv(T_(k))+v(T_(k+1)) and └y┘=Bv(q)+v(T_(k−1)). A left or right affine function (left or right move) is determined by the state q and the current tape square T_(k). If states q≠r, then |Bv(q)−Bv(r)|≧B. If two alphabet symbols a, b are distinct, then |v(a)−v(b)|<|A|.

Thus, if two distinct program instructions have different states q≠r, then the corresponding unit squares have y-coordinates that differ by at least B−|A|=|Q|≧2, since any Turing program with just one state has trivial behavior that can be determined in |A| execution steps when the tape is bounded. Otherwise, two distinct program instructions must have distinct symbols at T_(k). In this case, the corresponding unit squares have x-coordinates that differ by at least B−|A|=|Q|>2.

Definition 2.29 Rationally Bounded Coordinates

Let ƒ₁, ƒ₂, . . . , ƒ_(I) denote the I affine functions with corresponding unit square domains W₁, W₂, . . . , W_(I). Let p be a point in the plane P with orbit [p, ƒ_(S(1))(p), ƒ_(S(2))ƒ_(S(1))(p), . . . , ƒ_(S(m))ƒ_(S(m−1)) . . . ƒ_(S(2))ƒ_(S(1))(p), . . . ]. The orbit of p has rationally bounded coordinates if conditions I & II hold.

I) For every point in the orbit z=f_(S(k))ƒ_(S(k−1)) . . . ƒ_(S(2))ƒ_(S(1))(p) the x-coordinate of z, x(z), and the y-coordinate of z, y(z), are both rational numbers.

II) There exists a natural number M such that for every point

$z = \left( {\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}}} \right)$

in the orbit, where p₁, p₂, q₁, and q₂ are integers in reduced form, then |p₁|<M, |p₂|<M, |q₁|<M, and |q₂|<M.

An orbit is rationally unbounded if the orbit does not have rationally bounded coordinates.

Theorem 2.30 An Orbit with Rationally Bounded Coordinates is Periodic or Halting.

Proof. Suppose both coordinates are rationally bounded for the whole orbit and M is the natural number. If the orbit is halting we are done. Otherwise, the orbit is immortal. Since there are less than 2 M integer values for each one of p₁, p₂, q₁ and q₂ to hold, then the immortal orbit has to return to a point that it was already at. Thus it is periodic.

COROLLARY 2.31 A Turing machine execution whose tape head location is unbounded over the whole program execution corresponds to an immortal orbit.

THEOREM 2.32 Suppose the initial tape contents is bounded as defined in definition 2.2. Then an orbit with rationally unbounded coordinates is an immortal orbit that is not periodic.

PROOF. If the orbit halts, then the orbit has a finite number of points. Thus, it must be an immortal orbit. This orbit is not periodic because the coordinates are rationally unbounded.

COROLLARY 2.33 If the Turing Machine execution is unbounded on the right half of the tape, then in regard to the corresponding affine orbit, there is a subsequence S(1), S(2), . . . , S(k), . . . of the indices of the function sequence g₁, g₂, . . . , g_(k), . . . such that for each natural number n the composition of functions g_(S(n))g_(S(n−1)) . . . g_(S(1)) iterated up to the s(n)th orbit point is of the form

${\begin{pmatrix} B^{n} & 0 \\ 0 & \frac{1}{B^{n}} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m_{s{(n)}} \\ t_{s{(n)}} \end{pmatrix}$

where m_(s(n)), t_(s(n)) are rational numbers.

COROLLARY 2.34 If the Turing Machine execution is unbounded on the left half of the tape, then in regard to the corresponding affine orbit, there is a subsequence S(1), S(2), . . . , S(k), . . . of the indices of the function sequence g₁, g₂, . . . , g_(k), . . . such that for each natural number n the composition of functions g_(S(n))g_(S(n−1)) . . . g_(S(1)) iterated up to the s(n)th orbit point is of the form:

${\begin{pmatrix} \frac{1}{B^{n}} & 0 \\ 0 & B^{n} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}} + \begin{pmatrix} m_{s{(n)}} \\ t_{s{(n)}} \end{pmatrix}$

where m_(s(n)), t_(s(n)) are rational numbers.

Theorem 2.35 M-Bounded Execution Implies a Halting or Periodic Orbit

Suppose that the Turing Machine (Q, A, η) begins or continues execution with a configuration such that its tape head location is M-bounded during the next (2M+1)|Q∥A|^(2M+1)+1 execution steps. Then the Turing Machine program halts in at most (2M+1)|Q∥A|^(2M+1)+1 execution steps or its corresponding orbit is periodic with period less than or equal to (2M+1)|Q∥A|^(2M+1)+1.

PROOF. If the program halts in (2M+1)∥Q∥A|^(2M+1)+1 steps, then the proof is completed. Otherwise, consider the first (2M+1)|Q∥A|^(2M+1)+1 steps. There are a maximum of |Q∥A| program commands for each tape head location. There are a maximum of (2M+1) tape head locations. For each of the remaining 2M tape squares, each square can have at most |A| different symbols, which means a total of |A|^(2M) possibilities for these tape squares. Thus, in the first (2M+1)|Q∥A|^(2M+1)+1 points of the corresponding orbit in P, there are at most distinct (2M+1)|Q∥A|^(2M+1) points so at least one point in the orbit must occur more than once.

Prime Edge Complexity, Periodic Points & Repeating State Cycles

Definition 3.1 Overlap Matching & Intersection Patterns

The notion of an overlap match expresses how a part or all of one pattern may match part or all of another pattern. Let V and W be patterns. (V, s) overlap matches (W, t) if and only if V(s+c)=W(t+c) for each integer c satisfying λ≦c≦μ such that λ=min{s, t} and μ=min{|V|−1−s, |W|−1−t} where 0≦s<|V| and 0≦t<|W|. The index s is called the head of pattern V and t is called the head of pattern W. If V is also a subpattern, then (V, s) submatches (W, t).

If (V, s) overlap matches (W, t), then define the intersection pattern I with head u=λ as (I, u)=(V, s)∩(W, t), where I(c)=V(c+s−λ) for every integer c satisfying 0≦c≦(μ+λ) where λ=min {s, t} and μ=min {|V|−1−s, |W|−1−t}.

Definition 3.2 Edge Pattern Substitution Operator

Consider pattern V=v₀ v₁ . . . v_(n), pattern W=w₀ w₁ . . . w_(n) with heads s, t satisfying 0≦s, t≦n and pattern P=p₀ p₁ . . . p_(m) with head u satisfying 0≦u≦m. Suppose (P, u) overlap matches (V, s). Then define the edge pattern substitution operator ⊕ as E=(P, u)⊕[(v, s)

(W, t)] according to the four different cases A., B., C. and D.

Case A.) u>s and m−u>n−s. See FIG. 24.

${E(k)} = \begin{Bmatrix} {W\left( {k + s - u} \right)} & {when} & {u \leq {k + s} \leq {u + n}} \\ {P(k)} & {{{when}\mspace{14mu} 0} \leq k < {u - {s\mspace{14mu} {OR}}}} & {{u + n - s} < k \leq m} \end{Bmatrix}$

where the head of E is u+t−s. Observe that |E|=m+1

Case B.) u>s and m−u≦n−s. See FIG. 25.

${E(k)} = \begin{Bmatrix} {W\left( {k + s - u} \right)} & {when} & {{u - s} \leq k \leq {n + s - u}} \\ {P(k)} & {when} & {0 \leq k < {u - s}} \end{Bmatrix}$

where the head of E is u+t−s. Also, |E|=n+s−u+1

Case C.) u≦s and m−u<n−s. See FIG. 26.

E(k)=W(k) when 0≦k≦n and the head of E is t. Also, |E|=|W|=n+1.

Case D.) u≦s and m−u>n−s. See FIG. 27.

${E(k)} = \begin{Bmatrix} {W\left( {k + u - s} \right)} & {when} & {n < k \leq {m + s - u}} \\ {W(k)} & {when} & {0 \leq k \leq n} \end{Bmatrix}$

where the head of E is t. Also, |E|=m+s−u+1

Overlap and intersection matching and edge pattern substitution are useful in searching for periodic points and in algorithms that execute all possible finite Turing machine configurations.

Example 3.3 Overlap Matching and Edge Substitution

Set pattern P=0101 110. Set pattern V=11 0101. Set pattern W=01 0010. Then (P, 0) overlap matches (V, 2). Edge pattern substitution is well-defined so E=(P, 0)⊕[(V, 2)

(W, 4)]=01 0010 110. The head or index of pattern E=4. Also, (P, 4) overlap matches (V, 0). F=(P, 4)⊕[(V, 0)

(W, 4)]=0101 010010. The index of pattern F=u+t−s=4+4−0=8.

Definition 3.4 State Cycle

Consider N execution steps of Turing Machine (Q, A, η). After each execution step, the machine is in some state q_(k) and the tape head is pointing to some alphabet symbol a_(k). Relabeling the indices of the states and the alphabet symbols if necessary and assuming the machine has not halted after N execution steps in terms of the input commands is denoted as: (q₀, a₀)

(q₁, a₁).

. . .

(q_(N−1), a_(N−1))

(q_(N), a_(N)). A state cycle is a valid execution sequence of input commands such that the first and last input command in the sequence have the same state i.e. (q_(k), a_(k))

(q_(k+1), a_(k+1))

. . .

(q_(N−1), a_(N−1))

(q_(k), a_(k)). The length of this state cycle equals the number of input commands minus one. A state cycle is called a prime state cycle if it contains no proper state subcycles. For a prime state cycle, the length of the cycle equals the number of distinct states in the sequence. For example, (2, 0)

(3, 1)

(4, 0)

(2, 1) is called a prime 3-state cycle because it has length 3 and also 3 distinct states {2, 3, 4}.

Remark 3.5 any Prime State Cycle has Length ≦|Q|

This follows from the Dirichlet principle and the definition of a prime state cycle.

Remark 3.6 Maximum Number of Distinct Prime State Cycles

Given an alphabet A and states Q, consider an arbitrary prime state cycle with length 1, (q, a)

(q, b). There are |Q∥A| choices for the first input command and |A| choices for the second input command since the states must match. Thus, there are |Q∥A|² distinct prime state cycles with length 1. Similarly, consider a prime state cycle with window of execution whose length is 2, this can be represented as (q₁, a₁)

(q₂, a₂)

(q₁, b₁).

Then there are |Q∥A| choices for (q₁, a₁) and once (q₁, a₁) is chosen there is only one choice for q₂ because it is completely determined by η(q₁, a₁)=(q_(z), b₁) where n is the program in (Q, A, η). Similarly, there is only one choice for b₁. There are |A| choices for a₂. Thus, there are |Q∥A|² distinct choices.

For an arbitrary prime state cycle (q₁, a₁)

(q_(z), a₂)

. . .

(q_(n), a_(n))

(q₁, a_(n+1)) with window of execution of length k then there are |Q∥A| choices for (q₁, a₁) and |A| choices for a₂ since the current window of execution length after the first step increases by 1. There is only one choice for q₂ because it is determined by η(q₁, a₁). Similarly, for the jth computational step, if the current window of execution length increases by 1, then there are |A| choices for (q_(j+1), a_(j+1)). Similarly, for the jth computational step, if the current window of execution stays unchanged, then there is only one choice for a_(j+1) that was determined by one of the previous j computational steps. Thus, there are at most |Q∥A|^(k) distinct prime state cycles whose window of execution length equals k. Definitions 2.8 and remark 2.10 imply that a prime k-state cycle has a window of execution length less than or equal to k. Thus, from the previous and 3.5, there are at most

${Q}{\sum\limits_{k = 1}^{Q}{A}^{k}}$

distinct prime state cycles in (Q, A, η).

Remark 3.7 any State Cycle Contains a Prime State Cycle

PROOF. Relabeling if necessary let S(q₁, q₁)=(q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, a_(n+1)) be a state cycle. If q_(i) is the only state visited twice, then the proof is completed. Otherwise, define u=min{|S(q_(k), q_(k)) S(q_(k), q_(k)) is a subcycle of S(q₁, q₁)}. Then μ exists because S(q₁, q₁) is a subcycle of S(q₁, q₁). Claim: Any state cycle S(q_(j), q_(j)) with |S(q_(j), q_(j))|=μ must be a prime state cycle. Suppose not. Then there is a state r≠q_(j) that is visited twice in the state cycle S(q_(j), q_(j)). But then S(q_(r), q_(r)) is a cycle with length less than μ which contradicts μ's definition.

Definition 3.8 Consecutive Repeating State Cycle for (Q, A, η)

If machine (Q, A, η) starts execution and repeats a state cycle two consecutive times i.e. (q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁), then (Q, A, η) has a consecutive repeating state cycle.

Definition 3.9 Execution Node for (Q, A, η)

An execution node (or node) is a triplet Π=[q, w₀ . . . w_(n), t] for some state q in Q where w₀ w₁ . . . w_(n) is a pattern of n+1 alphabet symbols each in A such that t is a non-negative integer satisfying 0≦t≦n. Intuitively, w₀ w₁ . . . w_(n) is the pattern of alphabet symbols on n+1 consecutive tape squares on the tape and t represents the location of the tape head.

LEMMA 3.10 Every Immortal Periodic Point Induces a Consecutive Repeating State Cycle.

PROOF. Suppose p is an immortal periodic point with period n. Then by the Turing-Affine correspondence theorem the kth iterate of p is ƒ_(S(k)) ƒ_(S(k−1)) . . . ƒ_(S(1))(p) and the application of affine function ƒ_(S(k)) corresponds to the execution of input command (q_(k), b_(k)). Thus, let the input command sequence (q₁, b₁)

. . .

(q_(n), b_(n))

(q_(n+1), b_(n+1)) denote the first n input commands that are executed. Since p has period n, ƒ_(S(n)) . . . ƒ_(S(k)) . . . ƒ_(S(1))(p)=p. Thus, (q₁, b₁)=(q_(n+1), b₊₁). Thus, the first n steps are a state cycle (q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁). Since the n+1 computational step corresponds to applying ƒ_(S(1)) to p which corresponds to input command (q₁, b₁). By induction, the n+k computational step corresponds to applying function ƒ_(S(k)) to the point ƒ_(S(k−1)) . . . ƒ_(S(1))(p) which by the previous paragraph corresponds to the execution of the input command (q_(k), b_(k)). Thus, the sequence of input commands is (q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁).

Lemma 3.11 Every Consecutive Repeating State Cycle Induces an Immortal Periodic Orbit

Suppose Turing machine (Q, A, η) begins or resumes execution at some tape square and repeats a state cycle two consecutive times. Then (Q, A, η) has an immortal periodic point and this state cycle induces the immortal periodic point.

PROOF. Let the state cycle that is repeated two consecutive times be denoted as (q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁)

. . .

(q_(n), b_(n))

(q₁, b₁). Let s_(k) denote the tape square right before input command (q_(k), b_(k)) is executed the first time where 1≦k≦n. Let T_(k) denote the tape square right before input command (q_(k), b_(k)) is executed the second time where 1≦k≦n.

Thus, the window of execution for the first repetition of the state cycle, right before input command (q₁, b₁) is executed a second time, denoted I_(n)={s₁, s₂, . . . s_(k), s_(k+1) . . . s_(n), s_(n+1)} where s_(n+1)=t₁. The window of execution for the second repetition of the state cycle is J_(n)={t₁, t₂, . . . , t_(n), t_(n+1)} where t_(n+1)=t_(n)+t₁−s_(n).

Furthermore, observe that the window of execution for the computational steps 1 thru k is I_(k)={s₁, s₂, . . . s_(k), s_(k+1)} where the tape square s_(k+1) is indicated after input command (q_(k), b_(k)) is executed the first time. Also, observe that the window of execution for the computational steps n+1 thru n+k is J_(k)={t₁, t₂, . . . , t_(k), t_(k+1)} where the tape square T_(k+1) is indicated after the input command (q_(k), b_(k)) is executed the second time (in the second repeating cycle).

Next a useful notation represents the tape patterns for each computational step. Then the proof is completed using induction.

Let V₁ denote the tape pattern—which is the sequence of alphabet symbols in the tape squares over the window of execution I_(n)—right before input command (q₁, b₁) is executed the first time. Thus, V₁(s₁)=b₁. Let V_(k) denote the tape pattern—which is the sequence of alphabet symbols in the tape squares over the window of execution I_(n)—right before input command (q_(k), b_(k)) is executed the first time. Thus, V_(k)(s_(k))=b_(k).

Let W₁ denote the tape pattern—which is the sequence of alphabet symbols in the tape squares over the window of execution J_(n)—right before input command (q₁, b₁) is executed the second time. Thus, W₁(t₁)=b₁. Let W_(k) denote the tape pattern—which is the sequence of alphabet symbols in the tape squares over the window of execution J_(n)—right before input command (q_(k), b_(k)) is executed the second time. Thus, W_(k)(t_(k))=b_(k).

Using induction, it is shown that V₁ on window of execution I_(n) equals W₁ on window of execution J_(n). This completes the proof.

Since (q₁, b₁) is the input command before computational step 1 and (q₁, b₁) is the input command before computational step n+1, then V₁(s₁)=b₁=W₁(t₁). Thus, V₁ restricted to window of execution I₁ equals W₁ restricted to window of execution J₁.

From the definition, η(q₁, b₁)=η(q₂, a₁, x) for some a₁ in A and where x equals L or R. Note that L represents a left tape head move and R a right tape head move.

Case x=R. A right tape head move.

Then s₂=s₁+1, t₂=t₁+1 and V₁(s₂)=b₂=W₁(t₂). It has already been observed that V₁(s₁)=b₁=W₁(t₁). Thus, V₁ restricted to the window of execution I₂ equals W₁ restricted on the window of execution J₂ Furthermore, the tape head is at s₁ right before computational step 1 and input command (q₁, b₁) is executed; the tape head is at t₁ right before computational step n+1 and input command (q₁, b₁) is executed.

Also, V₂(s₁)=a₁=W₂(t₁) and V₂(s₂)=b₂=W₂(t₂). Thus, V₂ restricted to the window of execution I₂ equals W₂ restricted to the window of execution J₂ Furthermore, the tape head is at s₂ right before computational step 2 with input command (q₂, b₂) is executed; the tape head is at t₂ right before computational step n+2 with input command (q₂, b₂) is executed.

Case x=L. A left tape head move.

Then s₂=s₁−1, t₂=t₁−1 and V₁(s₂)=b₂=W₁(t₂). And V₁(s₁)=b₁=W₁(t₁). Thus, V₁ restricted to the window of execution I₂ equals W₁ restricted on the window of execution J₂. Furthermore, the tape head is at s₁ right before computational step 1 and input command (q₁, b₁) is executed; the tape head is at t₁ right before computational step n+1 and input command (q₁, b₁) is executed.

Also, V₂(s₁)=a₁=W₂(t₁) and V₂(s₂)=b₂=W₂(t₂). Thus, V₂ restricted to the window of execution I₂ equals W₂ restricted to the window of execution J₂ Furthermore, the tape head is at s₂ right before computational step 2 and input command (q₂, b₂) is executed; the tape head is at t₂ right before computational step n+2 and input command (q₂, b₂) is executed. This completes the base case of induction.

Induction Hypothesis. Suppose that for the 1, 2, . . . , k−1 computational steps and the corresponding n+1, n+2, . . . , n+k−1 steps that for every i with 1≦i≦k

V₁ restricted to the window of execution I_(i) equals W₁ restricted on the window of execution J_(i); V₂ restricted to the window of execution I_(i) equals W₂ restricted on the window of execution J_(i); and . . . V_(i) restricted to the window of execution I_(i) equals W_(i) restricted on the window of execution J_(i).

Furthermore, the tape head is at s_(i) right before computational step i and input command (q_(i), b_(i)) is executed; the tape head is at t_(i) right before computational step n+i and input command (q_(i), b_(i)) is executed.

Induction Step. Since (q_(k), b_(k)) is the input command before computational step k and before computational step n+k, then V_(k)(s_(k))=b_(k)=W_(k)(t_(k)).

From the definition, η(q_(k), b_(k))=η(q_(k+1), a_(k), x) for some a_(k) in A and x equals L or R. Note that L represents a left tape head move and R a right tape head move.

Case x=R. A right tape head move for computational steps k and n+k.

By the inductive hypothesis V_(k) restricted to window of execution I_(k) equals W_(k) restricted to window of execution J_(k) and the only change to the tape and tape head after executing η(q_(k), b_(k))=η(q_(k+1), a_(k), R) for the steps k and n+k is that V_(k+1)(s_(k))=a_(k)=W_(k+1)(t_(k)) and V_(k+1)(s_(k+1))=b_(k+1)=W_(k+1)(t_(k+1)) and that the tape heads move right to s_(k+1) and T_(k+1) respectively.

Thus, V_(k+1) restricted to the window of execution I_(k+1) equals W_(k+1) restricted on the window of execution J_(k+1). And for each j satisfying 1≦j≦k, then V_(j) restricted to the window of execution I_(k+1) equals W_(j) restricted on the window of execution J_(k+1).

Case x=L. A left tape head move for computational steps k and n+k.

By the inductive hypothesis V_(k) restricted to window of execution I_(k) equals W_(k) restricted to window of execution J_(k) and the only change to the tape and tape head after executing η(q_(k), b_(k))=η(q_(k+1), a_(k), L) for the steps k and n+k is that V_(k)+₁(s_(k))=a_(k)=W_(k)+₁(t_(k)) and V_(k+1)(S_(k+1))=b_(k)+₁=W_(k+1)(t_(k+1)) and that the tape heads move left to s_(k+1) and T_(k+1) respectively.

Thus, V_(k+1) restricted to the window of execution I_(k+1) equals W_(k+1) restricted on the window of execution J_(k+1). And for each j satisfying 1≦j≦k, then V_(j) restricted to the window of execution I_(k+1) equals W_(j) restricted on the window of execution J_(k+1).

Prime Directed Sequences & Periodic Point Search

Definition 4.1 Prime Directed Edge from Head and Tail Execution Nodes

A prime head execution node Δ=[q, v₀ v₁ . . . v_(n), s] and prime tail execution node=[r, w₀ w₁ . . . w_(n), t] are called a prime directed edge if and only if all of the following hold:

-   -   When Turing machine (Q, A, η) starts execution, it is in state         q; the tape head is located at tape square s. For each j         satisfying 0≦j≦n tape square j contains symbol v_(j). In other         words, the initial tape pattern is v₀ v₁ . . . v_(s) . . .         v_(n).     -   During the next N computational steps, state r is visited twice         and all other states in Q are visited at most once. In other         words, the corresponding sequence of input commands during the N         computational steps executed contains only one prime state         cycle.     -   After N computational steps, where 1≦N≦|Q|, the machine is in         state r. The tape head is located at tape square t. For each j         satisfying 0≦j≦n tape square j contains symbol w_(j). The tape         pattern after the N computational steps is w₀ w₁ . . . w _(t) .         . . w_(n).     -   The window of execution for these N computational steps is [0,         n].

A prime directed edge is denoted as Δ

Γ or [q, v₀ v₁ . . . v_(n), s]

[r, w₀ w₁ . . . t]. The number of computational steps N is denoted as |Δ

Γ|.

Definition 4.2 Prime Input Command Sequence

3.4 introduced input commands. If (q₁, a₁)

. . .

(q_(n), a_(n)) is an execution sequence of input commands for (Q, A, η), then (q₁, a₁)

. . .

(q_(n), a_(n)) is a prime input command sequence if q_(n) is visited twice and all other states in the sequence are visited once. In other words, a prime input command sequence contains exactly one prime state cycle.

Notation 4.3 Prime Input Command Sequence Notation

Using the same notation as lemma 3.11, let V₁ denote the initial tape pattern—which is the sequence of alphabet symbols in the tape squares over the window of execution of the prime input command sequence—right before the first input command (q₁, a₁) in the sequence is executed. And let s₁ denote the location of the tape head i.e. V₁(s₁)=a₁. Let V_(k) denote the tape pattern right before the kth input command (q_(k), a_(k)) in the sequence is executed and let s_(k) denote the location of the tape head i.e. V_(k)(s_(k))=a_(k).

Definition 4.4 Composition of Prime Input Command Sequences

Let (q₁, a₁)

. . .

(q_(n), a_(n)) and (r₁, b₁)

. . .

(r_(m), b_(m)) be prime input command sequences where V_(k) denotes the tape pattern right before the kth input command (q_(k), a_(k)) with tape head at s_(k) with respect to V_(k) and W_(k) denotes the tape pattern right before the kth input command (r_(k), b_(k)) with tape head at t_(k) with respect to W_(k).

Suppose (V_(n), s_(n)) overlap matches with (W₁, t₁) and q_(n)=r₁. Then (q_(n), a_(n))=(r₁, b₁). And the composition of these two prime input command sequences is defined as (q₁, a_(i))

. . .

(q_(n), a_(n))

(r₂, b₂)

. . .

(r_(m), b_(m)). The composition is undefined if (V_(n), s_(n)) and (W₁, t₁) do not overlap match or q_(n)≠r₁.

If (q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, b₁) is a prime state cycle, then it is also prime input command sequence. For simplicity in upcoming lemma 4.15, it is called a composition of one prime input command sequence.

The purpose of these next group of definitions is to show that any consecutive repeating state cycle is contained inside a composition of prime input command sequences. From lemmas 3.10 and 3.11, there is a one to one correspondence between a consecutive repeating state cycle and an immortal periodic point.

If this consecutive repeating state cycle is rotated, then it is still part of the same periodic orbit of the original periodic point. Next it is shown that there is a one to one correspondence between prime input command sequences and prime directed edges. Subsequently, it is explained how to link match prime directed edges. Then it is demonstrated how to find all prime directed edges for a particular Turing machine or a digital computer program. If a particular Turing machine (digital computer program) has any immortal periodic points, then it will have corresponding consecutive repeating state cycles which will be contained in an edge sequence of prime directed edges that are link matched.

Example 4.5 Directed Partition Method

Start with the finite sequence (0, 4, 2, 3, 4, 1, 3, 0, 1, 2, 0, 4, 2, 3, 4, 1, 3, 0, 1, 2).

Partition Steps (0 4 2 3 4 1 3 0 1 2 0 4 2 3 4 1 3 0 1 2) ( (0 4 2 3) 4 1 3 0 1 2 0 4 2 3 4 1 3 0 1 2) 4 lies in (0 4 2 3). 1st element found. ( (0 4 2 3) (4 1 3 0) 1 2 0 4 2 3 4 1 3 0 1 2) 1 lies in (4 1 3 0). 2nd element found. ( (0 4 2 3) (4 1 3 0) (1 2 0 4) 2 3 4 1 3 0 1 2) 2 lies in (1 2 0 4). 3rd element found. ( (0 4 2 3) (4 1 3 0) (1 2 0 4) (2 3 4 1) 3 0 1 2) 3 lies in (2 3 4 1). 4th element found. ( (0 4 2 3) (4 1 3 0) (1 2 0 4) (2 3 4 1) (3 0 1 2)) 0 lies in (0 4 2 3). 5th element found.

Definition 4.6 Tuples

A tuple is a finite sequence of objects denoted as (σ₁, σ₂, . . . σ_(m)). The length of the tuple is the number of objects in the sequence denoted as |(σ₁, σ₂, . . . , σ_(m))|=m. For our purposes, the objects of the tuple may be states, input commands or natural numbers. (3) is a tuple of length one. (1, 4, 5, 6) is a tuple of length four. Sometimes the commas will be omitted as in the previous example. (4 6 0 1 2 3) is a tuple of length six. The 4 is called the first object in tuple (4 6 0 1 2 3). 1 is called a member of tuple (4 6 0 1 2 3).

Definition 4.7 Tuple of Tuples

A tuple of tuples is of the form (w₁, w₂, . . . , w_(n)) where each w_(k) may have a different length. An example of a tuple of tuples is ((3), (1, 4, 5, 6), (4, 5, 6)). Sometimes the commas are omitted: ((0 8 2 3) (1 7 5 7) (5 5 6)).

Definition 4.8 Directed Partition of a Sequence

A directed partition is a tuple of tuples (w₁, w₂, . . . , w_(n)) that satisfies Rules A and B.

-   -   Rule A. No object a occurs in any element tuple w_(k) more than         once.     -   Rule B. If w_(k) and w_(k+1) are consecutive tuples, then the         first object in tuple w_(k+1) is a member of tuple W_(k).

Example 4.9 Directed Partition Examples

((0 8 2 3) (8 7 5 4) (5 0 6)) is an example of a directed partition.

((0 8 2 3) (8 7 5 4) (5 0 6)) is sometimes called a partition tuple.

(0 8 2 3) is the first element tuple. And the first object in this element tuple is 0.

Element tuple (8 0 5 7 0 3) violates Rule A because object 0 occurs twice.

((0 8 2 3) (1 7 5 4) (5 0 6)) violates Rule B since 1 is not a member of element tuple (0 8 2 3).

Definition 4.10 Consecutive Repeating Sequence and Extensions

A consecutive repeating sequence is a sequence (x₁, x₂, . . . , x_(n), . . . , x_(2n)) of length 2n for some positive integer n such that x_(k)=x_(n+k) for each k satisfying 1≦k≦n. An extension sequence is the same consecutive repeating sequence for the first 2n elements (x₁ . . . x_(n) . . . x_(2n) . . . X_(2n+m)) such that x_(k)=x_(2n+k) for each k satisfying 1≦k≦m.

A minimal extension sequence is an extension sequence (x₁, . . . x_(2n+m)) where m is the minimum positive number such that there is one element in x_(2n), x_(2n+1), . . . , x_(2n+m) that occurs more than once. Thus, x_(2n+k)=x_(2n+m) for some k satisfying 0≦k<m.

For example, the sequence S=(4 2 3 4 1 3 0 1 2 0 4 2 3 4 1 3 0 1 2 0) is a consecutive repeating sequence and S=(4 2 3 4 1 3 0 1 2 0 4 2 3 4 1 3 0 1 2 0 4 2 3 4 1) is an extension sequence. S contains consecutive repeating sequence S.

Definition 4.11 Directed Partition Extension with Last Tuple Satisfying Rule B

Suppose (x₁ . . . x_(n) . . . x_(2n), x_(2n+1), . . . x_(2n+m)) is an extension of consecutive repeating sequence (x₁ . . . x_(n) . . . X_(2n)). Then (w₁, w₂, . . . , w_(r)) is a directed partition extension if it is a directed partition of the extension: The last tuple w_(r) satisfies Rule B if x_(2n+m) is the last object in tuple w_(r) and x_(m+1) lies in tuple w_(r).

For example, the extension S=(4 2 3 4 1 3 0 1 2 0 4 2 3 4 1 3 0 1 2 0 4 2 3) has directed partition extension ((4 2 3) (4 1 3 0) (1 2 0 4) (2 3 4 1) (3 0 1 2) (0 4 2 3)) and the last tuple satisfies Rule B since 4 lies in (0 4 2 3)

Method 4.12 Directed Partition Method

Given a finite sequence (x₁ . . . x_(n)) of objects. Initialize element tuple w₁ to the empty tuple, ( ) Initialize partition tuple P to the empty tuple, ( ) For each element x_(k) in sequence (x₁ . . . x_(n)) { if x_(k) is a member of the current element tuple w_(r) { Append element tuple w_(r) to the end of partition tuple so that P = (w₁ . . . w_(r)) Initialize current element tuple w_(r+1) = (x_(k))  }  else update w_(r) by appending x_(k) to end of element tuple w_(r) } The final result is the current partition tuple P after element x_(n) is examined in the loop. Observe that the tail of elements from (x₁ . . . x_(n)) with no repeated elements will not lie in the last element tuple of the final result P.

Example 4.13 Directed Partition Method Implemented in newLISP, www.newlisp.org.

(define (add_object element_tuple object) (if (member object element_tuple) nil  (append element_tuple (list object)) )) (define (find_partition seq) (let ( (partition_tuple ′( ) )   (element_tuple ′( ) )   (test_add nil) ) (dolist (object seq) (set ′test_add (add_object element_tuple object) ) (if test_add (set ′element_tuple test_add) (begin (set ′partition_tuple (append partition_tuple (list element_tuple) ) ) (set ′element_tuple (list object) )  ) ) )  partition_tuple )) (set ′seq ′(4234130120 4234130120 4234) ) > (find_partition seq) ( (423) (4130) (1204) (2341) (3012) (0423) )

4 lies in the last tuple (0 4 2 3)

Lemma 4.14 Every Consecutive Repeating Sequence has an Extension Sequence with a Directed Partition Such that the Last Tuple Satisfies the Rule B Property.

PROOF. As defined in 4.10, extend consecutive repeating sequence (x₁, x₂ . . . x_(2n)) to the extension sequence (x₁, x₂ . . . x_(2n), x_(2n+1) . . . x_(2n+m)) such that m is the minimum positive number such that there is one element in x^(2n), x_(2n+1) . . . X_(2n+m) that occurs more than once. Thus, X_(2n+k)=x_(2n+m) for some k satisfying 0≦k<m.

Apply method 4.12 to S=(x₁, x₂ . . . x_(2n), x_(2n+1) . . . x_(2n+m)). Then the resulting partition tuple P extends at least until element x_(2n) and the last tuple in P satisfies rule B. If the partition tuple P is mapped back to the underlying sequence of elements, then it is an extension sequence since it reaches element x_(2n).

Lemma 4.15 any Consecutive Repeating State Cycle is Contained in a Composition of One or More Prime Input Command Sequences.

PROOF. Let σ=[(q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, a₁)

. . .

(q_(n), a_(n))] be a consecutive repeating cycle. Method 4.12 & 4.14 show that this sequence of consecutive repeating input commands may be extended to a minimal extension sequence: [(q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, a₁)

. . .

(q_(m), a_(m))]

For simplicity, let v_(k) denote input command (q_(k), a_(k)).

Apply method 4.12 to (v₁, . . . v_(n) v₁ . . . v_(n) v₁ . . . v_(m)) so that the result is the partition tuple P=(w₁, . . . w_(r)). Then the sequence of element tuples in P represent a composition of one or more prime input command sequences. Rules A and B imply that for consecutive tuples w_(k)=(v_(k(1)) v_(k(2)) . . . v_(k(m))) and w_(k+1)=(v_((k+1)(1)) v_((k+1)(2)) . . . v_((k+1)(m))), then (q_(k(1)), a_(k(1)))

(q_(k(2)), a_(k(2)))

. . .

(q_(k(m)), a_(k(m)))

(q_(k+1(1)), a_((k+1)(1))) is a prime input command sequence. And 4.14 implies that the last tuple w_(r) corresponds to a prime input command sequence and that the consecutive repeating state cycle is contained in the partition P mapped back to the sequence of input commands.

Definition 4.16 Finite Sequence Rotation

Let (x₀ x₁ . . . x_(n)) be a finite sequence. A k-rotation is the resulting sequence (x_(k) x_(k+1) . . . x_(n) x₀ X₁ . . . x_(k−1)). The 3-rotation of (8 7 3 4 5) is (3 4 5 8 7). When it does matter how many elements it has been rotated, it is called a sequence rotation.

Definition 4.17 Rotating a State-Symbol Cycle

Let (q₁, a₁)

. . .

(q_(n), a_(n))

(q₁, b₁) be a state cycle. This state cycle is called a state-symbol cycle if a₁=b₁. A rotation of this state-symbol cycle is the state cycle (q_(k), a_(k))

. . .

(q_(n), a_(n))

(q₁, a₁)

. . .

(q_(k), a_(k)) for some k satisfying 0≦k≦n. In this case, the state-symbol cycle has been rotated by k−1 steps.

Lemma 4.18 any Consecutive Repeating Rotated State Cycle Generated from a Consecutive Repeating State Cycle Induces the Same Immortal Periodic Orbit.

PROOF. Let p be the immortal periodic point induced by this consecutive repeating state cycle. Rotating this state cycle by k steps corresponds to iterating p by the next k corresponding affine functions.

LEMMA 4.19 Prime Directed Edges

Prime Input Command Sequences

Prime directed edges and prime input command sequences are in 1 to 1 correspondence. PROOF. (

) Let Δ

Γ be a prime directed edge where Δ=[q, v₀ v₁ . . . v_(n), s] and Γ=[r, w₀ w₁ . . . w_(n), t]. From the definition of a prime directed edge, over the next N computational steps some state r is visited twice, all other states in Q are visited at most once and there is a sequence of input commands (q, v_(s))

(q₁, a₁

. . . (r, a_(k)) . . .

(r, w_(t)) corresponding to these N steps. This is a prime input command sequence.

(

) Let (q₁, a₁)

. . .

(q_(n), a_(n)) be a prime input command sequence with N computational steps. Then q_(n) is visited twice and all other states in the sequence are visited only once. Let v₀ V₁ . . . v, be the initial tape pattern over the window of execution during the N computational steps. Now a₁=v_(s) for some s. Let w₀ w₁ . . . w_(n) be the final tape pattern over the window of execution as a result of these N steps. Then a_(n)=v, for some t. Thus, [q, v₀ v₁ . . . v_(n), s]

[r, w₀ w₁ . . . w_(n), t] is a prime directed edge.

Remark 4.20 Upper Bound for the Number of Prime Directed Edges

Each prime head node determines a unique prime directed edge so an upper bound for head nodes provides an upper bound for the number of distinct prime directed edges. Consider prime head node [q, V, s]. There are |Q| choices for the state q. Any pattern that represents the window of execution has length ≦|Q|+1. Furthermore, by the previous remark any pattern P such that (V, s) submatches (P, t) for some t, then the resultant pattern is the same since V spans the window of execution. Thus, |A|^(|Q|+1) is an upper bound for the number of different patterns V.

Lastly, there are two choices for s in a |Q|+1 length pattern because the maximum number of execution steps is |Q| i.e. the tape head move sequence is L^(|Q|) or R^(|Q|). Thus, |Q| is an upper bound for the number of choices for s unless |Q|=1. The following bound works in the trivial case that |Q|=1. Thus, there are at most |Q|²|A|^(|Q|+1) prime directed edges.

Example 4.21 3-State Machine Prime Directed Edges and Prime Input Command Sequences

Consider Turing Machine (Q, A, η). Q={2, 3, 4} and 1 is the halting state. A={0, 1}. η is specified as follows. η(2, 0)=(3, 1, L). η(2, 1)=(4, 0, L). η(3, 0)=(4, 1, R). η(3, 1)=(4, 0, R). η(4, 0)=(1, 0, R). η(4, 1)=(2, 0, R).

Prime Directed Edges Prime Input Command Sequences [2, 000, 1] 

 [2, 100, 2] (2, 0) 

 (3, 0) 

 (4, 1) 

 (2, 0) [2, 100, 2] 

 [2, 000, 2] (2, 0) 

 (3, 1) 

 (4, 1) 

 (2, 0) [2, 11, 1] 

 [2, 00, 1] (2, 1) 

 (4, 1) 

 (2, 0) [2, 001, 1] 

 [2, 101, 2] (2, 0) 

 (3, 0) 

 (4, 1) 

 (2, 1) [2, 101, 1] 

 [2, 001, 2] (2, 0) 

 (3, 1) 

 (4, 1) 

 (2, 1) [3, 010, 0] 

 [3, 101, 1] (3, 0) 

 (4, 1) 

 (2, 0) 

 (3, 0) [3, 110, 0] 

 [3, 001, 1] (3, 1) 

 (4, 1) 

 (2, 0) 

 (3, 0) [4, 10, 0] 

 [4, 11, 1] (4, 1) 

 (2, 0) 

 (3, 0) 

 (4, 1) [4, 11, 0] 

 [4, 00, 1] (4, 1) 

 (2, 1) 

 (4, 0)

There are 9 distinct prime state cycles. Observe that |Q|²|A|^(|Q|+1)=3²(4²) 144.

Observe that |Q|(|A|+|A|²)=2(2+4)=12.

The upper bound in 4.20 is not sharp. Although sharp upper bounds for the number of prime directed edges can be important, the sharpness of the upper bounds does not affect the speed of method 4.34 in finding prime directed edges for a particular Turing machine.

In what follows prime directed edges are link matched so that for a given Turing Machine a method for finding consecutive repeating state cycles is demonstrated. It is proved that this method will find immortal periodic points if they exist.

Definition 4.22 Halting Execution Node

Suppose [q, v₀ v₁ . . . v_(n), s] is an execution node and over the next |Q| computational steps a prime state cycle is not found. In other words, a prime directed edge is not generated. Then the Turing machine execution halted in |Q| or less steps. Let W be a pattern such that (W, t) submatches (V, s) and W spans the window of execution until execution halts. Define the halting node as H=[q, W, t].

Definition 4.23 Prime Directed Edge Complexity

Remark 4.20 provides an upper bound on the number of prime directed edges. Let={Δ₁

Γ₁, . . . , Δ_(k)

Γ_(k), . . . , Δ_(N)

r_(N)} denote the finite set of prime directed edges for machine (Q, A, η). Define the prime directed edge complexity of Turing machine (Q, A, η) to be the number of prime directed edges denoted as |P|.

Observe that any digital computer program also has a finite prime directed edge complexity. This follows from the fact that any digital computer program can be executed by a Turing machine.

Definition 4.24 Overlap Matching of a Node to a Prime Head Node Execution Node Π Overlap Matches Prime Head Node Δ if and Only if the Following Hold.

-   -   Π=[r, w₀ w₁ . . . w_(n), t] is an execution node satisfying         0≦t≦n     -   Δ=[q, v₀ v_(j) . . . v_(n), s] is a prime head node satisfying         0≦s≦m     -   State q=State r.     -   W denotes pattern w₀ w₁ . . . w_(n) and V denotes pattern v₀ v₁         . . . v_(m)     -   Pattern (W, t) overlap matches (V, s) as defined in definition         3.1.

Lemma 4.25 Overlap Matching Prime Head Nodes are Equal

If Δ=[q, P, u] and Δ_(k)=[q, V, s] are prime head nodes and they overlap match, then they are equal. (Distinct edges have prime head nodes that do not overlap match.)

Proof.

0≦u≦|Δ_(j)| and 0≦s≦|Δ_(k)|. Let (I, m)=(P, u)∩(V, s) where m=min{s, u}. Suppose the same machine begins execution on tape I with tape head at m in state q. If s=u and |Δ_(j)|=|Δ_(k)|, then the proof is complete.

Otherwise, s≠u or |Δ_(j)|≠|Δ_(k)| or both. Δ_(f) has a window of execution [0, |Δ_(j)|−1] and Δ_(k) has window of execution [0, |Δ_(k)|−1]. Let the ith step be the first time that the tape head exits finite tape I. This means the machine would execute the same machine instructions with respect to Δ_(j) and Δ_(k) up to the ith step, so on the ith step, Δ_(j) and Δ_(k) must execute the same instruction. Since it exits tape I at the ith step, this would imply that either pattern P or V are exited at the ith step. This contradicts either that [0, |Δ_(j)|−1] is the window of execution for Δ_(j) or [0, |Δ_(k)|−1] is the window of execution for Δ_(k).

Definition 4.26 Edge Node Substitution Operator Π⊕(Δ

Γ)

Let Δ

Γ be a prime directed edge with prime head node Δ=[q, v₀ v . . . s] and tail node Γ=[r, w₀ w₁ . . . w_(n), t]. If execution node Π=[q, p₀ p₁ . . . p_(m), u] overlap matches Δ, then the edge pattern substitution operator from 3.2 induces a new execution node Π⊕(Δ

Γ)=[r, (P,u)⊕[(V,s)

(W,t)], k] with head k=u+t−s if u>s and head k=t if u≦s such that 0≦s, t≦n and 0≦u≦m and patterns V=v₀ v₁ . . . v_(n) and W=w₀ w₁ . . . w_(n) and P=p₀ p₁ . . . p_(m).

Definition 4.27 Prime Directed Edge Sequence and Link Matching

A prime directed edge sequence is defined inductively. Each element is a coordinate pair with the first element being a prime directed edge and the second element is an execution node. Each element is abstractly expressed as (Δ_(k)

r_(k),

The first element of a prime directed edge sequence is (Δ₁

Γ₁, Π₁) where Π₁=Γ₁, and Δ₁

Γ₁ is some prime directed edge in P. For simplicity in this definition, the indices in P are relabeled if necessary so the first element has indices equal to 1. If Π₁ overlap matches some non-halting prime head node Δ₂, the second element of the prime directed edge sequence is (A₂

Γ₂, Π₂) where Π₂=Π₁⊕(Δ₂

Γ₂). This is called a link match step.

Otherwise, Π₁ overlap matches a halting node, then the prime directed edge sequence terminates. This is expressed as [(Δ₁

Γ₁, Γ₁), HALT]. In this case it is called a halting match step.

If the first k−1 steps are link match steps, then the prime directed edge sequence is denoted as [(Δ₁

Γ₁, Π₁), (Δ₂

Γ₂, Π₂), . . . , (Δ_(k)

Γ_(k), Π_(k))] where Π_(j) overlap matches prime head node Δ_(j+1) and Π_(j+1)=Π_(j)⊕(Δ_(j+1)

Γ_(j+1)) for each j satisfying 0≦j≦k.

Notation 4.28 Edge Sequence Notation E([p₁, p₂, . . . , p_(k)], k)

To avoid subscripts of a subscript, p_(k) and the subscript p_((j)) represent the same number. As defined in 4.27, P={Δ₁

Γ₁, . . . Δ_(k)

Γ_(k), . . . , Δ_(N)

r_(N)} denotes the set of all prime directed edges. E([p₁], 1) denotes the edge sequence [(Δ_(p(1))

Γ_(p(1)), Π_(p(1))] of length 1 where Π_(p(1))=Γ_(p(1)) and 1≦p₁≦|P|. Next E([p₁, p₂], 2) denotes the edge sequence [(Δ_(p(1))

Γ_(p(1)), Π_(p(1))), (Δ_(p(2))

Γ_(p(2)), Π_(p(2)))] of length 2 where Π_(p(2))=Π_(p(1))⊕(Δ_(p(2))

Γ_(p(2))) and 1≦p₁, p₂≦|P|.

In general, E([p₁, p₂, . . . , p_(k)], k) denotes the edge sequence of length k which is explicitly [(Δ_(p(1))

Γ_(p(1)), Π_(p(1))), (Δ_(p(2))

Γ_(p(2)), Π_(p(2))), . . . , (Δ_(p(k))

Γ_(p(k)), Π_(p(k)))] where Π_(p(j+1))=Π_(p(j))⊕(Δ_(p(j+1))

Γ_(p(j+1))) for each j satisfying 1≦j≦k−1 and 1≦p(j)≦|P|.

Definition 4.29 Edge Sequence Contains a Consecutive Repeating State Cycle

Lemma 4.19 implies that an edge sequence corresponds to a composition of prime input commands. The expression an edge sequence contains a consecutive repeating state cycle is used if the corresponding sequence of prime input commands contains a consecutive repeating state cycle.

THEOREM 4.30 Any consecutive repeating state cycle of (Q, A, η) is contained in an edge sequence of (Q, A, η).

PROOF. This follows immediately from definition 4.29 and lemmas 4.15 and 4.19.

Remark 4.31 Period of an Immortal Periodic Point Contained in Edge Sequence

If E([p₁, p₂, . . . , p_(r)], r) contains a consecutive repeating state cycle, then the corresponding immortal periodic point has period

$\leq {\frac{1}{2}{\sum\limits_{k = 1}^{r}{{\left. \Delta_{\; {p{(k)}}}\Rightarrow\Gamma_{p{(k)}} \right.}.}}}$

PROOF. This follows from lemma 3.11 that a consecutive repeating state cycle induces an immortal periodic point. The length of the state cycle equals the period of the periodic point. Further, the number of input commands corresponding to the number of computational steps equals |Δ_(p(k))

Γ_(p(k))| in directed edge Δ_(p(k))

Γ_(p(k)).

Method 4.32 Finding a Consecutive Repeating State Cycle in an Edge Sequence

Given an edge sequence whose corresponding prime input command sequence (q₀, a₀)

(q₁, a₁)

. . .

(q_(N), a_(N)) has length N

Set n=N/2 if N is even; otherwise, set n=(N+1)/2 if N is odd

for each k in {1, 2, . . . , n}

{ for each j in {0, 1, . . ., N−2k−1} {  if sequence (q_(j), a_(j)) 

 (q_(j+1), a_(j+1)) 

 . . .  

 (q_(j+k), a_(j+k)) equals  sequence (q_(j+k+1), a_(j+k+1)) 

 (q_(j+k+2), a_(j+k+2)) 

 . . .  

 (q_(j+2k+1),  a_(j+2k+1)) then { return consecutive repeating state cycle  (q_(j), a_(j)) 

 (q_(j+1), a_(j+1)) 

 . . .  

 (q_(j+k), a_(j+k)) 

 . . .  

 (q_(j+k+1),  a_(j+k+1)) }  } }

If exited outer for loop without finding a consecutive repeating state cycle

Return NO consecutive repeating state cycles were found.

Example 4.33 a newLISP Function that Finds a Consecutive Repeating Sequence

(define (find_pattern_repeats p_length seq) (let (  (k 0)  (max_k (− (length seq) (+ p_length p_length)) )  (pattern nil)  (repeat_pair nil)  (no_repeats true) ) (while (and (<= k max_k) no_repeats) (set ′pattern (slice seq k p_length)) (if (= pattern (slice seq (+ k p_length) p_length)) (begin (set ′repeat_pair (list pattern k)) (set ′no_repeats false) ) ) (set ′k (+ k 1)) )  repeat_pair )) (define (find_repeats seq) (let  ( (p_length 1) (max_p_length (/ (length seq) 2) ) (repeat_pair nil)  ) (while (and (<= p_length max_p_length) (not repeat_pair) )  (set ′repeat_pair (find_pattern_repeats p_length seq))  (set ′p_length (+ p_length 1)) ) repeat_pair )) (set ′s1 ′(3 5 7 2 3 5 7 11 5 7 )) ;; s1 does not have a consecutive repeating sequence. (set ′s2 ′(3 5 7 2 3 5 7 11 5 7 11 2 4 6 8 )) ;; 5 7 11 5 7 11 is a consecutive repeating sequence starting at element in list s2 (set ′s3 ′(1 2 0 2 1 0 2 0 1 2 0 2 1 0 1 2 1 0 2 1 2 0 2 1 0 1 2 0 2 1 2 0 1 2 1 0 1 2 0 1 0 1)) ;; 0 1 0 1 is a consecutive repeating sequence starting at element 38 in list s3 > (find_repeats s1) nil > (find_repeats s2) ( (5 7 11) 5) > (find_repeats s3) ( (0 1) 38)

Method 4.34 Prime Directed Edge Search Method

Given Turing Machine (Q, A, η) as input, the search method works as follows.

Set P=Ø.

For each non-halting state q in Q

For each pattern a_(−|Q| . . .) a⁻² a⁻¹ a₀ a₁ a₂ . . . a_(|Q|) selected from A^(2|Q|+1)

{

With tape head located at a₀, start executing machine (Q, A, η) until one state has been visited twice or (Q, A, η) reaches a halting state. The Dirichlet principle implies this will take at most |Q| computational steps. If it does not halt, let r be the state that is first visited twice. As defined in 4.1, over this window of execution, a prime directed edge Δ

 Γ is constructed where Δ = [q, v₀ v₁ . . . v_(n), s], Γ = [r, w₀ w₁ . . . w_(n), t] and 0 ≦ s, t ≦ n ≦ |Q|. Set P = P ∪ {Δ

 Γ} }

Remark 4.35 Prime Directed Edge Search Method Finds all Prime Directed Edges

Method 4.34 finds all prime directed edges of (Q, A, η) and all halting nodes.

PROOF. Let Δ

Γ be a prime directed edge of (Q, A, η). Then Δ

Γ has a head node A=[r, v₀ v₁ . . . v_(n), s], for some state r in Q, for some tape pattern v₀ v₁ . . . v_(n) that lies in A^(n+1), such that n≦|Q| and 0≦s≦n. In the outer loop of 4.34, when r is selected from Q and in the inner loop when the tape pattern a_(−|Q| . . .) a⁻² a⁻¹ a₀ a₁ a₂ . . . a_(|Q|) is selected from A^(2|Q|+1) such that

$\begin{matrix} {a_{0} = v_{s}} & {a_{1} = v_{s + 1}} & \ldots & {a_{k} = v_{s + k}} & \ldots & {a_{n - s} = v_{n}} \\ {a_{- 1} = v_{s - 1}} & {a_{- 2} = v_{s - 2}} & \ldots & {a_{- k} = v_{s - k}} & \ldots & {a_{- s} = v_{0}} \end{matrix}$

then the machine execution in 4.34 will construct prime directed edge Δ

Γ. When the head node is a halting node, the machine execution must halt in at most |Q| steps. Otherwise, it would visit a non-halting state twice and thus, be a non-halting head node. The rest of the argument for this halting node is the same as for the non-halting head node.

Method 4.36 Immortal Periodic Point Search Method

Given Turing Machine (Q, A, η) as input, the method works as follows.

Use method 4.34 to find all prime directed edges, P.

set k = 1. set Φ(1) = { E([1], 1), E([2], 1), . . ., E([|P|], 1) } while ( Φ(k) ≠ Ø ) { set Φ(k+1) = Ø. for each E([p₁,p₂, . . .,p_(k)], k) in Φ(k) { for each prime directed edge Δ_(j) 

 Γ_(j) in P { if Δ_(j) 

 Γ_(j) link matches with Π_(p(k)) then { set p_(k+1) = j set Φ(k+1) = Φ(k+1) ∪ E([p₁, p₂, . . ., p_(k),p_(k+1)], k+1) if E([p₁, p₂, . . ., p_(k), p_(k+1)], k+1) contains a consecutive repeating state cycle then return the consecutive repeating state cycle  }  } } k is incremented. } if (while loop exited because Φ(m)=Ø for some m) then return Ø

REMARK 4.37 |Φ(k)| is finite and |Φ(k)|≦|P|^(k)

PROOF. |Φ(1)|=|P|. Analyzing the nested loops, in method 4.36

for each E([p₁, p₂, . . . , p_(k)], k) in Φ(k)

-   -   for each Δ_(j)         Γ_(j) in P { . . . }

For each edge sequence E([p₁, p₂, . . . , p_(k)], k) chosen from Φ(k), at most |P| new edge sequences are put in Φ(k+1). Thus |Φ(k+1)|≦|P∥Φ(k)|, so |Φ(k)|≦|P|^(k).

Definition 4.38 Periodic Turing Machine

A Turing machine (Q, A, η) that has at least one periodic configuration, whenever it has an immortal configuration is said to be a periodic Turing machine.

THEOREM 4.39 If (Q, A, η) is a periodic machine, then method 4.36 terminates in a finite number of steps with either a consecutive repeating state cycle or for some positive integer J, then Φ(J)=Ø

PROOF. If (Q, A, η) has at least one configuration (q, k, T) that has an immortal orbit, then the assumption that (Q, A, η) is a periodic machine implies the existence of a periodic point p with some finite period N. Thus, from lemma 3.10, there is a consecutive repeating state cycle that corresponds to the immortal periodic orbit of p. Since method 4.36 searches through all possible prime edge sequences of length k, a consecutive repeating state cycle will be found that is contained in a prime directed edge sequence with length at most 2N. Thus, this immortal periodic point of period N will be reached before or while computing Φ(2N).

Otherwise, (Q, A, η) does not have any configurations with an immortal orbit; in other words, for every configuration, (Q, A, η) halts in a finite number of steps.

Claim: There is a positive integer J such that every edge sequence terminates while executing method 4.36. By reductio absurdum, suppose not. Then there is at least one infinite prime directed edge sequence that exists: this corresponds to an immortal orbit, which contradicts that (Q, A, η) does not have any configuration with an immortal orbit.

Periodic Point Properties

LEMMA 5.1 All Right Head Moves or all Left Tape Head Moves

If Turing Machine (Q, A, η) can execute |Q| consecutive computational steps which are all right tape head moves or all left tape head moves without halting, then the machine has an immortal periodic point with period ≦|Q|.

PROOF. Consecutive Right Tape Head Moves.

Suppose there are |Q| consecutive computational steps which are all right tape head moves. Thus, there is a sequence of |Q| commands η(q_(k), a_(k))=(q_(k−1), b_(k), R) satisfying 1≦k≦|Q| that execute these |Q| computational steps without halting. In the table below, the subscript k in each state q_(k) indicates the state of the machine just before the kth computational step.

. . .

b₂ . . . b_(k) . . .

k + |Q|

. . .

. . .

a _(|Q|+1) q₂ b₁ a ₂ . . . a_(k) . . . a_(|Q|) . . . q_(k) b₁ b₂ . . . a _(k) . . . a_(|Q|)

indicates data missing or illegible when filed

The Dirichlet Box principle applied to the |Q|+1 states {q₁, q₂ . . . q_(|Q|), q_(|Q|+1)} implies that two of the states are equal; namely, q_(j)=q_(k) for some j<k. Thus, the point

$p = \left\lbrack {q_{j},{\overset{\_}{b_{j}b_{{j + 1}\mspace{11mu}}\ldots \mspace{14mu} b_{k - 1}}\mspace{25mu} \overset{\_}{{\langle a_{j}\rangle}a_{j + 1}\mspace{14mu} \ldots \mspace{14mu} a_{k - 1}}}} \right\rbrack$

is an immortal periodic point with period k−j and (k−j)≦|Q|.

Consecutive Left Tape Head Moves. Suppose there are |Q| consecutive computational steps which are all left tape head moves. Thus, there is a sequence of |Q| commands η(q_(k), a_(k))=(q_(k+1), b_(k), L) satisfying 1≦k≦|Q| that execute these |Q| computational steps without halting. In the table below, the subscript k in each state q_(k) indicates the state of the machine just before the kth computational step.

STATE k − |Q| k − |Q| + 1 k q₁ a_(|Q|) . . . a_(k) . . . a₂ a ₁ q₂ a_(|Q|) . . . a_(k) . . . a ₂ b₁ . . . q_(k) a_(|Q|) . . . a _(k) b₂ b₁ . . . q_(|Q|) a _(|Q|) . . . b_(k) . . . b₂ b₁ q_(|Q|+1) a _(|Q|+1) b_(|Q|) . . . b_(k) . . . b2 b₁

The Dirichlet Box principle applied to the |Q|+1 states {q₁, q₂, . . . q_(|Q|), q_(|Q|+1)} implies that two of the states are equal; namely, q_(j)=q_(k) for some j<k. Thus, the point

$p = \left\lbrack {q_{j},{\overset{\_}{a_{k - 1}\mspace{14mu} \ldots \mspace{14mu} {\langle a_{j}\rangle}}\mspace{14mu} \overset{\_}{b_{{k - 1}\mspace{14mu}}\ldots \mspace{14mu} b_{j}}}} \right\rbrack$

is an immortal periodic point with period k−j and (k−j)≦|Q|.

Remark 5.2

Consider Turing Machine (Q, A, η). If for some q in Q, and for some a, b in A, η(q, a)=(q, b, R) or η(q, a)=(q, b, L), then (Q, A, η) has an immortal fixed point.

PROOF. If η(q, a)=(q, b, R), then p=[q, b

a

ā] is an immortal fixed point of (Q, A, η). If η(q, a)=(q, b, L), then p=[q, ā

a

b] is an immortal fixed point of (Q, A, η).

Definition 5.3 Minimal Period

If (Q, A, η) has no immortal orbits, then it is called a halting machine. Otherwise, the minimum{C(p): p is an immortal periodic point of (Q, A, η)} is well-defined because it is the minimum of a non-empty subset of the natural numbers. This minimum is called the minimal period of (Q, A, η).

Theorem 5.4 Two State Minimal Period Theorem

If Q={q, r} has only two non-halting states and A={0, 1}, then (Q, A, η) is a halting machine or its minimal period is 1, 2 or 4.

PROOF. If (Q, A, η) is a halting machine, then the proof is completed. Otherwise; suppose (Q, A, η) has an immortal periodic point with period ≧5. Then it is shown that this implies the machine must have an immortal periodic point with period ≦4.

Consider the first 5 computational steps of the immortal periodic point p. If two consecutive steps stay in the same state, then by remark 5.2, then (Q, A, η) has an immortal periodic point with period 1. Done. Furthermore, if there are two consecutive right tape head moves or left tape head moves, during these five computational steps, then by lemma 5.1, there is an immortal periodic point with period ≦2.

Thus, W.L.O.G. (symmetry), for the remaining cases the first five computational steps look like—where the variables x₁, x₂, x₃, x₄, y₁, y₂, y₃ represent elements of A:

STATE MOVE STEP q x ₁ y₁ r x₂ y ₁ R 1 q x ₂ y₂ L 2 r x₃ y ₂ R 3 q x ₃ y₃ L 4 r x₄ y ₃ R 5

Observation 1.

x₁=x₂ implies x₁=x₂=x₃ because computational steps 1 and 3 are η(q, x₁)=(r, x₂, R) and η(q, x₂)=(r, x₃, R).

Observation 2.

y₁=y₂ implies that y₁=y₂=y₃ because computational steps 2 and 4 are η(r, y₁)=(q, y₂, L) and η(r, y₂)=(q, y₃, L).

Observation 3.

Since A has 2 elements, [x₁=x₂ or x₁=x₃ or x₂=x₃] and [y₁=y₂ or y₁=y₃ or y₂=y₃].

CASE 1: x₂=x₃ and y₂=y₃. Based on steps 3 and 4, point p=[q, 0

x₂

y₂ 6] is an immortal periodic point with period 2.

Many of the nine cases below are reduced to previous cases.

CASE 2: x₁=x₂ and y₂=y₃. The first observation reduces case 2 to case 1.

CASE 3: x₁=x₃ and y₂=y₃. By replacing all occurrences of x₃ by x₁ and all occurrences of y₃ by y₂, then the previous table becomes:

STATE MOVE STEP q x ₁ y₁ r x₂ y ₁ R 1 q x ₂ y₂ L 2 r x₁ y ₂ R 3 q x ₁ y₂ L 4 r x₂ y ₂ R 5 q x ₂ y₂ L 6

After the substitution, from step 4, then η(r, y₂)=(q, y₂, L). This implies step 6 in the table.

Looking at steps 2 and 6, point p=[q, 0

x₂

y₂ 0] is an immortal periodic point with period 2 or 4.

CASE 4: x₁=x₃ and y₁=y₃. Substituting x₁ for x₃ and y₁ for y₃ in step 4 of the original table, then the point p=[q, 0

x₁

y₁ 0] is an immortal periodic point with period 2 or 4.

CASE 5: x₁=x₂ and y₁=y₃. This reduces to case 4 from the first observation.

CASE 6: x₂=x₃ and y₁=y₃. Substituting x₂ for x₃ and y₁ for y₃ in the original table and observing that from step 3 that η(q, x₂)=(q, x₂, R). This implies that x₄=x₂

STATE MOVE STEP q x ₁ y₁ r x₂ y ₁ R 1 q x ₂ y₂ L 2 r x₂ y ₂ R 3 q x ₂ y₁ L 4 r x₂ y ₁ R 5

Then observe that after step 1 and step 5, the points are identical. Thus, the point p=[r, 0x₂

y₁

0] is immortal with period equal to 2 or 4.

CASE 7: x₁=x₂ and y₁=y₂.

This reduces to case 2 from the second observation.

CASE 8: x₁=x₃ and y₁=y₂. This reduces to case 4 from the second observation.

CASE 9: x₂=x₃ and y₁=y₂. This reduces to case 6 from the second observation.

Finally, it is shown that any machine having an immortal periodic point with period 3 must have an immortal periodic point with period 1 or 2. Suppose the machine has an immortal period 3 point. During the three computational steps, the claim is that there has to be two consecutive steps that are in the same state. For example, the state sequence q, r, q, r would contradict that it is a period 3 orbit because at step 0 it is in state q and after step 3 it is in state r; similarly, the state sequence r, q, r, q would contradict that is a period 3 orbit. Thus, there must be two consecutive steps that are in the same state, which implies it is an immortal fixed point. Thus, the machine can not have a minimal period of 3.

Immortal Orbit and Recurrent Points

In this section, recurrent points (configurations) are defined and the Immortal Orbit theorem is shown, which generalizes the behavior in the two previous examples. In particular, if machine (Q, A, η) has an immortal configuration (q, k, T), then the corresponding orbit of the point p=φ(q, k, T) with respect to the corresponding affine functions has a recurrent point and at least one recurrent point z is a fixed point of a composition sequence of these affine functions iterated according to the correspondence in Theorem 2.22

Definition 6.1 Recurrent Points

Let X be a metric space, S: N→N a function index sequence, and a family of functions ƒ_(k): X→X, where kεN. The orbit of point pεX is {p, ƒ_(S(1))(p), ƒ_(S(2))ƒ_(S(1))(p), . . . , ƒ_(S(m))ƒ_(S(m−1)) . . . ƒ_(S(2)) ƒ_(S(1))(p), . . . }. The ω-limit set, ω(p, ƒ_(k)), is the set of cluster points of the orbit of p i.e.

${\omega \left( {p,f_{k}} \right)} = {\bigcap\limits_{n \in N}{\overset{\_}{\left\{ {{f_{S{(k)}} \circ f_{S{({k - 1})}} \circ \mspace{14mu} \ldots \mspace{14mu} \circ {f_{1}(p)}}:{k > n}} \right\}}.}}$

Any point zεX that lies in ω(p, ƒ_(k)) is called a recurrent point. The closure of the set of recurrent points of ƒ_(k) is denoted R(ƒ_(k)) and is called the recurrent set of ƒ_(k).

An aperiodic Turing machine has at least one immortal configuration, but has no periodic immortal configurations. This first example is an aperiodic Turing machine.

Example 6.2 Aperiodic Turing Machine Execution and Affine Orbit Correspondence

Consider the Turing machine with state set Q={a, b, c, d, e, f}, halting state h and alphabet A={0, 1, 2, 3}. The Turing program η is shown below.

η(a, 0) = (d, 1, R) η(a, 1) = (f, 1, R) η(a, 2) = (f, 2, R) η(a, 3) = (f, 3, R) η(b, 0) = (c, 1, L) η(b, 1) = (e, 1, L) η(b, 2) = (e, 2, L) η(b, 3) = (e, 3, L) η(c, 0) = (a, 2, L) η(c, 1) = (f, 1, R) η(c, 2) = (f, 2, R) η(c, 3) = (f, 3, R) η(d, 0) = (b, 2, R) η(d, 1) = (e, 1, L) η(d, 2) = (e, 2, L) η(d, 3) = (e, 3, L) η(e, 0) = (a, 3, L) η(e, 1) = (a, 0, L) η(e, 3) = (d, 0, R) η(e, 3) = (f, 0, R) n(f, 0) = (b, 3, R) η(f, 1) = (b, 0, R) η(f, 2) = (c, 0, L) η(f, 3) = (e, 0, L)

Base B=|Q|A|+1=11. The value function v: A∪Q∪{h}→N is defined explicitly as v(h)=0. v(0)=1. v(1)=2. v(2)=3. v(3)=4. v(a)=5. v(b)=6. v(c)=7. v(d)=8. v(e)=9. and v(f)=10. Before program execution starts, we assume the machine is in state b. See FIGS. 28 and 29. The tape head is at tape square 2. T(1)=2, T(0)=1 and all other tape squares contain 0 i.e. the initial machine configuration is [b, 0 0 0]. The first computational step is η(b, 0)=(c, 1, −1) which is equivalent to applying the affine map

${f_{1}\left( {x,y} \right)} = {\left( {{{\frac{1}{11}x} + 34},{{11y} - 682}} \right)\mspace{14mu} {to}\mspace{14mu} {the}\mspace{14mu} {point}\mspace{14mu} {\left( {{12\frac{1}{10}},{69\frac{21}{110}}} \right).}}$

The calculation of the x-coordinate is

${{11{v(0)}} + {v(0)} + \frac{\frac{1}{11}}{1 - \frac{1}{11}}} = {12{\frac{1}{10}.}}$

The y-coordinate is

${{11{v(b)}} + {v(2)} + {\frac{1}{11}{v(1)}} + \frac{\frac{1}{11^{2}}}{1 - \frac{1}{11}}} = {69{\frac{21}{110}.}}$

Observe that the machine configuration [b, 0 0 0] is a recurrent point of this orbit starting at point [b, 0120 0]. Furthermore,

${\phi \left\lbrack {b,{\overset{\_}{0}\underset{\_}{0}\overset{\_}{0}}} \right\rbrack}\mspace{11mu} {\left( {{12\frac{1}{10}},{67\frac{1}{10}}} \right).}$

Observe that

${f_{3}f_{2}{f_{1}\left( {x,y} \right)}} = \left( {{{11x} - 121},{{\frac{1}{11}y} + 61}} \right)$

has fixed point

$\left( {{12\frac{1}{10}},{67\frac{1}{10}}} \right)$

which corresponds to recurrent point [b, 0 0 0]. Also, observe that

${f_{12}f_{11}\mspace{14mu} \ldots \mspace{14mu} f_{5}{f_{4}\left( {x,y} \right)}} = {\left( {{{11x} - 121},{{\frac{1}{11}y} + 61}} \right).}$

In other words, this recurrent point of the immortal orbit is a fixed point of a finite composition sequence of the corresponding affine maps.

A Turing machine that has at least one periodic configuration, whenever it has an immortal configuration is called a periodic Turing machine.

EXAMPLE 6.3 Q={q₀, q₁ q₂} A={#, a, b}

η(q₀, #)=(q₁, a, R). η(q₀, a)=(q₀, a, L). η(q₀, b)=(q₀, b, L).

η(q₁, #)=(q₀, b, L). η(q₁, a)=(q₁, a, R). η(q₁, b)=(q₁, b, R).

η(q₂, #)=(h, #, R). η(q₂, a)=(h, a, R). η(q₂, b)=(h, b, R).

The initial tape contents are all #'s. Formally, T(k)=# for every integer k. Base B=|Q|+|A|+1=7. The value function v: A∪Q∪{h}→N is explicitly v(h)=0 v (#)=1 v (a)=2 v(b)=3 v(q₀)=4 v(q₁)=5 v(q₂)=6.

In regard to the affine orbit and Turing Machine execution correspondence see FIGS. 30, 31 and 32. The affine functions ƒ_(j) and unit square domains W_(j), the current Turing Machine state, tape head and tape are shown after each execution step. The location of the tape head is indicated by underlining the tape symbol and placing the state of the machine directly underneath this tape square.

The point

$u = {\left( {{14 + 3 + \frac{\frac{3}{7}}{1 - \frac{1}{7}}},{35 + 2 + \frac{\frac{2}{7}}{1 - \frac{1}{7}}}} \right) = \left( {{17\frac{1}{2}},{37\frac{1}{3}}} \right)}$

is a recurrent point that corresponds to the machine being in state q₁ and the tape head location is at square 1 and the tape contents are T(k)=a whenever k≦1 and T(k)=b whenever k>1.

The point

$v = {\left( {{{2 \cdot 7} + 2 + \frac{\frac{2}{7}}{1 - \frac{1}{7}}},{{5 \cdot 7} + 2 + \frac{\frac{2}{7}}{1 - \frac{1}{7}}}} \right) = \left( {{16\frac{1}{3}},{37\frac{1}{3}}} \right)}$

is a recurrent point of the immortal orbit that corresponds to the machine being in state q₁ and the tape head location can be at any square j and the tape contents are T(k)=a for every k on the tape. Observe that v is a fixed point of

${f_{14}\left( {x,y} \right)} = \left( {{{7\; x} - 98},{{\frac{1}{7}y} + 32}} \right)$

lying in the interior of the unit square W₁₄=[(16, 37), (17, 37), (17, 38), (16, 38)].

Set

$v_{0} = \left( {{16\frac{145}{294}},{37\frac{1}{6}}} \right)$

which is when the tape head is at 0; the machine is in state q₁; and T(0−1)=T(0)=T(0+1)=a. Then v₀ is in this immortal orbit.

Tape Coordinate −1 0 1 2 3 4 5 6 7 8 Tape a a a b b # # # # # Contents State q₁

Set v₁=to the first time the point in the immortal orbit has its tape head at 4; and the machine is in state q₀; and T(−1−2)=T(−1−1)=T(−1)=T(−1+1)=T(−1+2)=a.

Tape Coordinate −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 Tape # a a a a a b b b b # # # Contents State q₁

By induction, for every k>1, set v_(k)=to the first time the point in the immortal orbit has its tape head at −k; the machine is in state q₀; and T(−k+j)=a for every j such that |j|≦k+1. This implies that

${\lim\limits_{n\rightarrow\infty}v_{n}} = {v.}$

The point

$w = {\left( {\frac{21 \cdot 7}{7 - 1},\frac{- 189}{1 - 7}} \right) = \left( {{24\frac{1}{2}},{31\frac{1}{2}}} \right)}$

is a recurrent point of the immortal orbit that corresponds to the machine being in state q₀ and the tape head location can be at any square j and the tape contents are T(k)=b for every k on the tape. Observe that w is a fixed point of η₈(x,y)=( 1/7x+21, 7y−189) lying in the interior of the unit square W₈=[(24, 31), (25, 31), (25, 32), (24, 32)]. Also, observe that if

$w_{3} = \left( {{24\frac{1}{6}},{31\frac{685}{2085}}} \right)$

which is when the tape head is at 3; and the machine is in state q₀; and T(3−1)=T(3)=T(3+1)=b, then w₃ is in this immortal orbit.

Tape Coordinate −1 0 1 2 3 4 5 6 7 8 Tape a a a b b b # # # # Contents State q₀

Set w₄ equal to the first time the point in the immortal orbit has its tape head at 4; and the machine is in state q₀; and T(4−2)=T(4−1)=T(4)=T(4+1)=T(4+2)=b.

Tape Coordinate −4 −3 −2 −1 0 1 2 3 4 5 6 1 8 Tape # a a a a a b b b b b # # Contents State q₀

By induction, for every k>4, set w_(k) equal to the first time the point in the immortal orbit has its tape head at k; and the machine is in state q₀; and T(k+j)=b for every j such that |j|≦k−2. Observe that

${\lim\limits_{n\rightarrow\infty}w_{n}} = {w.}$

The next theorem states what was already illustrated in the previous examples.

Theorem 6.4 an Immortal Orbit has Recurrent Points.

PROOF. Let ε>0. Let ƒ₁, ƒ₂, . . . , ƒ₁ denote the I affine functions with corresponding unit square domains W₁, W₂, W₃, . . . , W_(I). Let p be an immortal point. Then its orbit

$\left\{ {p,{f_{S{(1)}}(p)},{f_{S{(2)}}{f_{S{(1)}}(p)}},\ldots \mspace{14mu},{f_{S{(m)}}f_{S{({m - 1})}}\mspace{14mu} \ldots \mspace{14mu} f_{S{(2)}}{f_{S{(1)}}(p)}},\ldots}\mspace{14mu} \right\} \subseteq {\overset{I}{\bigcup\limits_{k = 1}}{W_{k}.}}$

The area of

$\overset{I}{\bigcup\limits_{k = 1}}W_{k}$

equals I. There exists a natural number N such that

$\frac{1}{N} < {\frac{ɛ}{\sqrt{2}}.}$

Consider the first I(N²+1) points of this orbit. Then at least N²+1 points of the orbit must lie in at least one of these unit squares W_(k). This implies that there are two points from the orbit in unit square W_(k) with the distance between them less than or equal to

$\frac{1}{N}.$

By choosing each N_(k) such that

$N_{k} < {\frac{1}{2}N_{k - 1}}$

and using induction, create a subsequence {p₁, p₂, . . . } of the immortal orbit of p such that p_(k)→z. Then z is a recurrent point.

Definition 6.5 Cluster Point

Let {p_(n)}_(n=1) ^(∞) be a sequence of points and d a metric. The point q is a cluster point of {p_(n)}_(n=1) ^(∞) if for any ε>0 and for any N, there exists m≧N such that d(q, p_(m))<ε.

Definition 6.6 Immortal Points

Let

${U\left( {Q,A,\eta} \right)} = \left\{ {p \in {\overset{I}{\bigcup\limits_{k = 1}}{W_{k}:}}} \right.$

p has an immortal orbit with respect to machine (Q, A, η)}U is called the set of immortal points with respect to Turing machine (Q, A, η).

Theorem 6.7 Immortal Points are a Closed Set

Suppose

${\lim\limits_{n\rightarrow\infty}p_{n}} = p$

and for each n that p_(n) lies in U(Q, A, η). Then p is an immortal point.

PROOF. Since

${\lim\limits_{n\rightarrow\infty}p_{n}} = p$

a subsequence q_(n) of p_(n) may be constructed with the property that

${d\left( {q_{n},p} \right)} < \frac{1}{B^{n}}$

where base B=|Q|+|A|+1. This means the corresponding tape of p and the corresponding tape of q_(n) are the same for the first n symbols on the tape in both directions.

p Tape Contents . . . b_(n+1) b_(n) . . . b₁ a ₀ a₁ . . . a_(n) a_(n+1) . . . State with tape w head at a ₀ q_(n) Tape Contents . . . d_(n+1) b_(n) . . . b₁ a ₀ a₁ . . . a_(n) c_(n+1) . . . State with tape w head at a ₀

By reductio absurdum, suppose p is not an immortal point. Then execution of (Q, A, η) on p halts at the Nth execution step for some natural number N. Since each computational step of (Q, A, η) moves either one tape square to the left or to the right, this means that execution of (Q, A, η) on p must halt on some tape symbol a_(k) or b_(k) such that k≦N. Assuming execution begins at tape square 0 or renumbering if necessary: this means that (Q, A, η) starting at point p must halt before moving to the right of tape square N or to the left of tape square −N. Consider point q_(N+2). By assumption, q_(N+2) is immortal but the first N+2 execution steps of (Q, A, η) on q_(N+2) and p must be the same because their tape symbols agree in both directions for at least the first N+2 tape squares. This is a contradiction: p halts on execution step N and q_(N+2) does not halt on step N. Thus, p must be an immortal point.

LEMMA 6.8 Every point in an immortal orbit and all cluster points lie in the interior of

$\bigcup\limits_{k = 1}^{I}{W_{k}.}$

Each point is a distance at least

$\frac{1}{B - 1}$

from the boundary of every W_(k).

PROOF. Consider a non-halting Turing machine execution. After every execution step, the state in the machine configuration is not in a halting state, h and the tape's range is in A. Further, recall that v(h)=0, v(a₁)=1, v(a_(|A|))=|A| and v(q_(|Q|))=B−1. If the tape contents of every tape square is a₁, this corresponds to x and y coordinates whose fractional parts are

$\frac{\frac{1}{B}}{1 - \frac{1}{B}} = {\frac{1}{B - 1}.}$

If the tape contents of every tape square is a_(|A|), which is the tape symbol with the largest value with respect to v, then this corresponds to x and y coordinates whose fractional parts are

$\frac{\frac{A}{B}}{1 - \frac{1}{B}} = {\frac{A}{B - 1}.}$

Thus, the point p in the immortal orbit corresponding to each configuration has the property that

${{{{x(p)} - \left\lfloor {x(p)} \right\rfloor}} \geq \frac{1}{B - 1}},{{{{y(p)} - \left\lfloor {y(p)} \right\rfloor}} \geq \frac{1}{B - 1}}$ ${{{\left\lceil {x(p)} \right\rceil - {x(p)}}} \geq {\frac{\left( {B - 1} \right)}{B - 1} - \frac{A}{B - 1}}} = \frac{Q}{B - 1}$ and ${{{\left\lceil {y(p)} \right\rceil - {y(p)}}} \geq {\frac{\left( {B - 1} \right)}{B - 1} - \frac{A}{B - 1}}} = {\frac{Q}{B - 1}.}$

Thus, any cluster point of the immortal orbit must be at least

$\frac{1}{B - 1}$

from the boundary of each W_(k).

Before the main result is reached, it is helpful to do some elementary counting of finite tape patterns (strings) based on the finite alphabet A and the finite state set Q regardless of the particular Turing machine program

Example 6.9 The alphabet A={1, 2, 3}

s₁s₂=12 is a particular 2-string where s₁=1 and s₂=2

The 2-string 12 occurs four times on the above tape. The 2-string 12 is non-overlapping. The 3-string s₁s₂ s₃=123 occurs two times on the above tape.

Definition 6.10 Alphabet Strings

Suppose alphabet A={a₁, a₂, a₃ . . . , a_(n)}. A k-string is a string of k symbols denoted as s₁s₂ s₃ . . . s_(k) where each s_(j) lies in A.

Definition 6.11 Overlapping and Disjoint Tape Strings

Let T_(k) denote the alphabet symbol on the kth square of the Turing tape. Then T_(k) T_(k+1) . . . T_(n), where k is an n−k+1 string of alphabet symbols. The two strings T_(j) T_(j+1) . . . T_(m) and T_(k) T_(k+1) . . . T_(n) are disjoint if m<k OR n<j. Otherwise, they are called overlapping strings.

Furthermore, overlapping strings are distinct if j≠k OR m≠n.

Definition 6.12 Identical Tape Strings

The two tape strings T_(j) T_(j+1) . . . T_(m) and T_(k) T_(k+1) . . . T_(n) are identical if they are the same length (i.e. n−k=m−j) and T_(k+i)=T_(j+i) for each i satisfying 1≦i≦n−k.

Definition 6.13 Identical Program Execution up to B^(n)

Two Turing machines M₁(Q₁, A₁, η_(i)) and M₂(Q₂, A₂, η₂) have identical program execution up to B^(n) if there is an isomorphism Ψ: M₁→M₂ such that the following conditions hold.

-   -   (1) Both φ: Q₁→Q₂ and γ: A₁→A₂ are bijective functions.     -   (2) Ψ(η₁)=η₂ such that Ψη₁(q, a)=η₂(φ(q), γ(a))     -   (3) Machine M₁ has tape head location j and machine M₂ has tape         head location k.     -   (4) Ψ(T_(j+m))=Γ_(k+m) for every m such that −n≦m≦n where T is         the tape for M₁ and Γ is the tape for M₂.     -   (5) M₁ program execution is in state r and M₂ program execution         is in state q.     -   (6) φ(r)=q.

The same machine can be identical to itself up to B^(n) at two different steps in its program execution.

Example 6.14 Overlapping 3-Strings

Consider A={0, 1} and string 000 001 010 011 100 101 110 111 a₁a₂a₃. The substring at the end, a₁a₂a₃, repeats one of the previous 3-strings. In a string of length |A|³+3 at least one 3-string is repeated. For example, in the following string of length 11 where a₁a₂a₃ . . . a₁₁=0101 0011 101, observe that a₂a₃a₄=a₉a₁₀a₁₁=101.

Example 6.15 Distinct Overlapping 2-Strings

In any string of length |A|²+2, two distinct overlapping 2-strings are repeated.

$\begin{matrix} {A = \left\{ {0,1} \right\}} & 0000 \\ \; & 0001 \\ \; & 00101 \\ \; & 00100 \\ \; & 001101 \\ \; & 01000 \\ \; & 01001 \\ \; & 0101 \\ \; & 011000 \\ \; & 011001 \\ \; & 0111 \\ \; & \ldots \end{matrix}$

Duality of 0 and 1 for the remaining eight have corresponding repeats as the first eight.

REMARK 6.16 Distinct overlapping n-strings that are substrings of any string of length L=|A|^(n)+n implies that at least two substrings are identical.

PROOF. Let a₁ a₂ a₃ . . . a_(z), be any alphabet string constructed from A. Define the set of n-strings S={a_(j+1) a_(j+2) . . . a_(j+n): 0≦j≦|A|^(n)}. All elements of S are substrings of a₁a₂a₃ . . . a_(L). By the definition of S, there are |A|^(n)+1 distinct overlapping substrings of a₁a₂a₃ . . . a_(L) but the maximum number of non-identical n-strings constructed from A is |A|^(n). Thus, at least two n-strings must be identical.

LEMMA 6.17 Distinct overlapping n-strings that are substrings of any string of length L=|A|^(n)+n implies that at least m+1 substrings are identical.

PROOF. Define S={a_(j+1) a_(j+2) . . . a_(j+n): 0≦j≦m|A|^(n)}. The rest of the proof is similar to the proof in 6.16.

COROLLARY 6.18 Any Turing tape string of length L=|Q∥A|^(n)+n has at least one n-string that occurs |Q|+1 times. In other words, the tape string has |Q|+1 identical, distinct, overlapping substrings.

PROOF. This follows immediately from lemma 6.17 and definitions 6.10, 6.11 and 6.12.

Definition 6.19 Composition Sequence of Affine Maps

Consider Turing machine (Q, A, η). Suppose that over the next n computational steps, the sequence of machine configurations is (q₀, k₀, T₀), (q₁, k_(n), T₁), . . . , (q_(n), k_(n), T_(n)). Let p=φ(q₀, k₀, T₀).

According to Theorem 2.22, consider corresponding orbit, [p, ƒ_(S(1))(p), ƒ_(S(2))∘ƒ_(S(1))(p), . . . , ƒ_(S(n))∘ƒ_(S(n−1)) . . . ƒ_(S(2)) ƒ_(S(1)))(p), . . . ]. Then for each k≧1 and for every m≧k, the affine map ƒ_(S(m))∘ƒ_(S(m−1))∘ . . . ∘ƒ_(S(k)) is called a composition sequence of affine functions or a composition sequence.

THEOREM 6.20 If machine (Q, A, η) has an immortal configuration (q, k, T), then Theorem 6.4 implies the orbit of the point p=φ(q, k, T) with respect to the corresponding affine functions has at least one recurrent point. Moreover, there is at least one recurrent point z of this immortal orbit that is a fixed point of a composition sequence of the affine functions iterated according to the correspondence in 2.22

PROOF. Let ƒ₁, ƒ₂, . . . ƒ_(I) denote I affine functions with corresponding unit square domains W₁, W₂, W₃, . . . , W_(I) induced by a Turing machine (Q, A, η) where base B=|Q|+|A|+1. Let p=(x(q, k, T), y(q, k, T)) for some immortal configuration (q, k, T) with respect to (Q, A, η). Then the orbit of p, [p, ƒ_(S(1))(p), ƒ_(S(2))ƒ_(S(1))(p), . . . , ƒ_(S(m))ƒ_(S(m−1)) . . . ƒ_(S(2)) ƒ_(S(1))p, . . . ], is an immortal orbit.

For convenience of notation, set p₀=p, p₁=f_(S(1))(p), p₂=ƒ_(S(2))ƒ_(S(1))(p), . . . , p_(m)=ƒ_(S(m))ƒ_(S(m−1)) . . . ƒ_(S(2))ƒ_(S(1))(p), . . . Define the set Ω={p_(k): k≧0} i.e. the immortal orbit considered as a set instead of a sequence. If the immortal orbit [p₀, P₁, p_(z), . . . , p_(m), . . . ] is periodic, then we are done as we have an immortal periodic point. Otherwise, set Ω contains an infinite number of distinct points. From 6.4 using induction, we can construct a subsequence {s_(k)} of the immortal orbit such that

${d\left( {s_{k},s_{k + 1}} \right)} \leq \frac{1}{k^{2}}$

for every k. It follows that

${\lim\limits_{n->\infty}s_{n}} = {z.}$

Thus, z is a cluster point of subsequence {s_(k)} which is a subset of Ω. And theorem 6.7 implies that z is an immortal point.

By choosing appropriate elements from the subsequence {s_(k)} that converges to z, there exists a subsequence of the immortal orbit [p₀, p₁, p₂, . . . , p_(m), . . . ] with the following properties:

A.

${\lim\limits_{n->\infty}z_{n}} = z$

B. For all n,

$z_{n} \in {{interior}\bigcup\limits_{k = 1}^{I}{W_{k}\mspace{14mu} {and}\mspace{14mu} z}} \in {{interior}\mspace{14mu} \bigcup\limits_{k = 1}^{I}{W_{k}.}}$

C.

${d\left( {z_{n},z} \right)} < \frac{1}{B_{n}}$

D. z_(n+1)=r_(n)(z_(n)) where r_(n) is the affine function that is the composition of the finite sequence of functions mapping z_(n) to z_(n+1) in the immortal orbit.

Consider the sequence of squares [a_(n), b_(n), c_(n), d_(n)] with center z_(n) and sides with length

$\frac{4}{B^{n - 1}}$

for n>1. See FIG. 30. From lemma 6.8, z is in the interior of W_(j) for some j. Set δ=infimum{d(p, z): pε∂W}. Since z is in the interior, then δ>0. Thus, there is a natural number N such that

$\frac{4\sqrt{2}}{B^{N - 1}} < {\frac{\delta}{2}.}$

Thus, for n≧N this geometrically means that if z is in the square [a_(n), b_(n), c_(n), d_(n)] then this whole square is a subset of W_(j). Also, observe that

$\begin{matrix} {{d\left( {z_{n},z_{n + 1}} \right)} \leq {{d\left( {z_{n},z} \right)} + {d\left( {z,z_{n + 1}} \right)}}} \\ {< {\frac{1}{B^{n}} + \frac{1}{B^{n + 1}}}} \\ {< \frac{2}{B^{n}}} \\ {\leq {\frac{2}{N}\frac{1}{B^{n - 1}}}} \\ {\leq \frac{1}{B^{n - 1}}} \end{matrix}$ because B ≥ 2.

See FIG. 31. The previous inequalities imply that z_(n+1) and z are in the interior of square [a_(n), b_(n), c_(n), d_(n)] since z_(n) is at the center. Thus, square [a_(n), b_(n), c_(n), d_(n)] is a subset of W_(j) for n≧N.

Covering Condition.

CLAIM: If for some n≧N, the linear part of at least one r_(n) is of the form

$\begin{pmatrix} B^{m} & 0 \\ 0 & \frac{1}{B^{m}} \end{pmatrix}\mspace{14mu} {or}\mspace{14mu} \begin{pmatrix} \frac{1}{B^{m}} & 0 \\ 0 & B^{m} \end{pmatrix}$

for some m≧1, then r_(n)([a_(n), b_(n), c_(n), d_(n)]) horizontally or vertically covers the square [a_(n), b_(n), c_(n), d_(n)].

PROOF: In the case that r_(n) is of the form

$\begin{pmatrix} \frac{1}{B^{m}} & 0 \\ 0 & B^{m} \end{pmatrix},$

recall that y(p) denotes the y coordinate of point p and x(p) denotes the x coordinate of point p.

Condition A. in Definition 1.8

Thus,

${{y\left( {r_{n}\left( c_{n} \right)} \right)} - {y\left( {r_{n}\left( a_{n} \right)} \right)}} = {{B^{m}\frac{4}{B^{n - 1}}} \geq \frac{4B}{B^{n - 1}}}$

where B≧2 and m≧1.

Observe that

${y\left( z_{n + 1} \right)} = {\frac{1}{2}{\left\{ {{y\left( {r_{n}\left( a_{n} \right)} \right)} + {y\left( {r_{n}\left( c_{n} \right)} \right)}} \right\}.}}$

Thus,

${{y\left( {r_{n}\left( c_{n} \right)} \right)} - {y\left( z_{n + 1} \right)}} = {{\frac{1}{2}\left\{ {{y\left( {r_{n}\left( c_{n} \right)} \right)} - {y\left( {r_{n}\left( a_{n} \right)} \right)}} \right\}} \geq \frac{4}{B^{n - 1}}}$

From above

${d\left( {z_{n},z_{n + 1}} \right)} \leq \frac{1}{B^{n - 1}}$

implies that y (r_(n)(a_(n)))<y(a_(n)) because

${{y\left( c_{n} \right)} - {y\left( a_{n} \right)}} = {\frac{4}{B^{n - 1}}.}$

The inequality y(r_(n)(a_(n)))<y(a_(n)) satisfies condition A in definition 1.8.

Condition B. in Definition 1.8

Similarly,

${{y\left( {r_{n}\left( d_{n} \right)} \right)} - {y\left( {r_{n}\left( b_{n} \right)} \right)}} \geq \frac{4B}{B^{n - 1}}$

where B≧2 and m≧1.

Observe that

${y\left( z_{n + 1} \right)}\frac{1}{2}{\left\{ {{y\left( {r_{n}\left( d_{n} \right)} \right)} + {y\left( {r_{n}\left( b_{n} \right)} \right)}} \right\}.}$

Thus,

${{y\left( {r_{n}\left( d_{n} \right)} \right)} - {y\left( z_{n + 1} \right)}} = {{\frac{1}{2}\left\{ {{y\left( {r_{n}\left( d_{n} \right)} \right)} - {y\left( {r_{n}\left( b_{n} \right)} \right)}} \right\rbrack} \geq \frac{4}{B^{n - 1}}}$

From above

${d\left( {z_{n},z_{n + 1}} \right)} \leq \frac{1}{B^{n - 1}}$

implies y(r_(n)(d_(n)))>y(d_(n)) because

${{y\left( d_{n} \right)} - {y\left( b_{n} \right)}} = \frac{4}{B^{n - 1}}$

The inequality y(r_(n)(d_(n)))>y(d_(n)) satisfies condition B in definition 1.8.

Condition C. in Definition 1.8

Observe that

${{x\left( {r_{n}\left( b_{n} \right)} \right)} - {x\left( {r_{n}\left( a_{n} \right)} \right)}} = {{\frac{1}{B^{m}}\frac{4}{B^{n - 1}}} \leq \frac{2}{B^{n - 1}}}$

where B≧2 and m≧1.

And since

${x\left( z_{n + 1} \right)} = {{{\frac{1}{2}\left\{ {{x\left( {r_{n}\left( a_{n} \right)} \right)} + {x\left( {r_{n}\left( b_{n} \right)} \right)}} \right\} \mspace{14mu} {then}\mspace{14mu} {x\left( z_{n + 1} \right)}} - {x\left( {r_{n}\left( a_{n} \right)} \right)}} \leq {\frac{1}{B^{n - 1}}.}}$

Thus,

$\begin{matrix} {{{{x\left( {r_{n}\left( a_{n} \right)} \right)} - {x\left( z_{n} \right)}}} \leq {{{{x\left( {r_{n}\left( a_{n} \right)} \right)} - {x\left( z_{n + 1} \right)}}} + {{{x\left( z_{n + 1} \right)} - {x\left( z_{n} \right)}}}}} \\ {\leq {\frac{1}{B^{n - 1}} + \frac{1}{B^{n - 1}}}} \\ {= \frac{2}{B^{n - 1}}} \end{matrix}$

Since

${{x\left( z_{n} \right)} - {x\left( a_{n} \right)}} = \frac{2}{B^{n - 1}}$

then x(r_(n)(a_(n)))≧x(a_(n)) which satisfies condition C in 1.8

Condition D. in Definition 1.8

Observe that

${{x\left( {r_{n}\left( b_{n} \right)} \right)} - {x\left( {r_{n}\left( a_{n} \right)} \right)}} = {{\frac{1}{B^{m}}\frac{4}{B^{n - 1}}} \leq \frac{2}{B^{n - 1}}}$

where B≧2 and m≧1.

And since

${x\left( z_{n + 1} \right)} = {{{\frac{1}{2}\left\{ {{x\left( {r_{n}\left( a_{n} \right)} \right)} + {x\left( {r_{n}\left( b_{n} \right)} \right)}} \right\} \mspace{14mu} {then}\mspace{14mu} {x\left( {r_{n}\left( b_{n} \right)} \right)}} - {x\left( z_{n + 1} \right)}} \leq {\frac{1}{B^{n - 1}}.}}$

Thus,

$\begin{matrix} {{{{x\left( {r_{n}\left( b_{n} \right)} \right)} - {x\left( z_{n} \right)}}} \leq {{{{x\left( {r_{n}\left( b_{n} \right)} \right)} - {x\left( z_{n + 1} \right)}}} + {{{x\left( z_{n + 1} \right)} - {x\left( z_{n} \right)}}}}} \\ {\leq {\frac{1}{B^{n - 1}} + \frac{1}{B^{n - 1}}}} \\ {= {\frac{2}{B^{n - 1}}.}} \end{matrix}$

Since

${{x\left( b_{n} \right)} - {x\left( z_{n} \right)}} = \frac{2}{B^{n - 1}}$

then x(r_(n)(b_(n)))≦x(b_(n)) which satisfies condition D in 1.8.

In the case where r_(n) is of the form

$\begin{pmatrix} B^{m} & 0 \\ 0 & \frac{1}{B^{m}} \end{pmatrix},$

then conditions A., B., C., and D. of definition 1.10 are verified in a similar manner. Thus, for this particular n≧N, then r_(n) has a fixed point in square [a_(n), b_(n), c_(n), d_(n)]⊂W_(j). By theorems 1.13 and 1.14, the vertical and horizontal covering fixed point theorems respectively, r_(n) has a fixed point which is an immortal periodic point.

Otherwise, for some M, for all n≧M, the linear part of r_(n) is the identity

$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$

Set U=maximum{M, N} where N was chosen above such that

$\frac{4\sqrt{2}}{B^{N - 1}} < {\frac{\delta}{2}.}$

Set L=≡1Q∥A|^((2U+1))+(2U+1). If there exists n≧L with r_(n)(x, y)=(x, y), the proof is completed.

Otherwise, from 2.26 and 2.29, the program execution is unbounded on the left of the tape or the right of the tape or both. W.L.O.G. it is assumed that program execution is unbounded on the right. (The argument for unbounded on the left is similar.) Thus, when n≧L, then z_(n) corresponds to the tape head being located at some position k, the tape contents are fixed at least n symbols in both directions from the tape square k and the machine is in some state q₁. In other words, z_(n) corresponds to

Tape Location k − k − k − k − k + k + k + n . . . 3 2 1 k 1 2 . . . n Tape b_(n) . . . b₃ b₂ b₁ a₀ a₁ a₂ . . . a_(n) . . . Contents State q₁ (Tape head)

Since the program execution is unbounded on the right, for the next net L moves to the right on the tape, consider the state q when the tape head is located at tape square k+U+j for each j such that 0≦j≦L. By lemma 6.14, there are |Q|+1 distinct tape (2U+1)-strings that are identical where the tape head lies at a distinct tape location that is in the middle of this (2 U+1)-string. Since there are |Q| states, this means there are at least two of these strings that are identical and are also in the same state. Thus, there is an affine map r from the one identical (2U+1)-string (point v) to the other identical (2U+1)-string (point w); in other words, r(v)=w where map r's linear part is hyperbolic

$\quad\begin{pmatrix} B^{m} & 0 \\ 0 & \frac{1}{B^{m}} \end{pmatrix}$

such that m≧1. Let [a_(u), b_(v), c_(v), d_(v)] be the square with center v and side length equal to

$\frac{4}{B^{U}}.$

Thus, r(v)=w and

${d\left( {v,w} \right)} < \frac{1}{B^{U}}$

implies that r([a_(v), b_(v), c_(v), d_(v)]) horizontally covers the square [a_(v), b_(v), c_(v), d_(v)]. By 1.14, r has a fixed point in [a_(v), b_(v), c_(v), d_(v)] which is an immortal periodic point.

The work in the previous proof yields a sharper result. In particular, if m, n≧L, then z_(m) and z_(n) are identical up to B^(L). Thus, the execution of the machine for the next net L tape head moves to the right are identical. Thus, the immortal orbit [p, ƒ_(S(1))(p), . . . , ƒ_(S(m)) ƒ_(S(m−1)) . . . ƒ_(S(2)) ƒ_(S(1))(p), . . . ] has a subsequence that converges to a fixed point in square [a_(v), b_(v), c_(v), d_(v)] with center v.

REMARK 6.21 By making each of the (2U+1)-strings in theorem 6.20 at least m tape squares apart the horizontal covering rectangle or vertical covering rectangle can be made thinner than ε>0 where the fixed point lies inside the covering rectangle.

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TABLE 1 Boolean Functions f_(k): {0, 1} × {0, 1} → {0, 1} Function AEM Separation Rule(s) f₁ =  

$\left. \left( {0,0,{- 1},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {0,0} \right),\left( {1,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}}{0/} \right.$ f₂ = ⊥ $\left. \left( {0,0,1,X_{k}} \right)\leftrightarrow\frac{0/}{\left\{ {\left( {0,0} \right),\left( {1,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}} \right.$ f₃ = A $\left. \left( {2,0,1,X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {1,1} \right),\left( {1,0} \right)} \right\}}{\left\{ {\left( {0,1} \right),\left( {0,0} \right)} \right\}} \right.$ f₄ = B $\left. \left( {0,2,1,X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {1,1} \right),\left( {0,1} \right)} \right\}}{\left\{ {\left( {1,0} \right),\left( {0,0} \right)} \right\}} \right.$ f₅ =  

 A $\left. \left( {{- 2},0,{- 1},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {0,1} \right),\left( {0,0} \right)} \right\}}{\left\{ {\left( {1,1} \right),\left( {1,0} \right)} \right\}} \right.$ f₆ =  

 B $\left. \left( {0,{- 2},{- 1},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {1,0} \right),\left( {0,0} \right)} \right\}}{\left\{ {\left( {1,1} \right),\left( {0,1} \right)} \right\}} \right.$ f₇ = A  

 B $\left. \left( {2,2,3,X_{k}} \right)\leftrightarrow\frac{\left\{ \left( {1,1} \right) \right\}}{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {0,0} \right)} \right\}} \right.$ f₈ = A  

 B $\left. \left( {2,2,1,X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}}{\left\{ \left( {0,0} \right) \right\}} \right.$ f₉ = A → B $\left. \left( {{- 4},2,{- 3},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {0,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}}{\left\{ \left( {1,0} \right) \right\}} \right.$ f₁₀ = A ← B $\left. \left( {2,{- 4},{- 3},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {0,0} \right),\left( {1,0} \right),\left( {1,1} \right)} \right\}}{\left\{ \left( {0,1} \right) \right\}} \right.$ f₁₁ = A  

 B (2, −4, −3, X_(k)) and (−4, 2, −3, X_(j)) with j ≠ k f₁₂ =  

 (A  

 B) $\left. \left( {{- 2},{- 2},{- 1},X_{k}} \right)\leftrightarrow\frac{\left\{ \left( {0,0} \right) \right\}}{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}} \right.$ f₁₃ =  

 (A  

 B) $\left. \left( {{- 2},{- 2},{- 3},X_{k}} \right)\leftrightarrow\frac{\left\{ {\left( {1,0} \right),\left( {0,1} \right),\left( {0,0} \right)} \right\}}{\left\{ \left( {1,1} \right) \right\}} \right.$ f₁₄ = A ⊕ B (2, 2, 1, X_(k)) and (−2, −2, −3, X_(j)) with j ≠ k f₁₅ = A < B $\left. \left( {{- 2},4,3,X_{k}} \right)\leftrightarrow\frac{\left\{ \left( {0,1} \right) \right\}}{\left\{ {\left( {0,0} \right),\left( {1,0} \right),\left( {1,1} \right)} \right\}} \right.$ f₁₆ = A > B $\left. \left( {4,{- 2},3,X_{k}} \right)\leftrightarrow\frac{\left\{ \left( {1,0} \right) \right\}}{\left\{ {\left( {0,0} \right),\left( {0,1} \right),\left( {1,1} \right)} \right\}} \right.$

TABLE 2 Minsky Universal Turing Machine with Program η y 0 1 A q₁ (q₁, 0, L) (q₁, 0, L) (q₂, 1, L) (q₁, 1, L) q₂ (q₁, 0, L) (q₂, y, R) (q₂, A, R) (q₆, y, R) q₃ (q₃, y, L) (h, 0, h) (q₃, A, L) (q₄, 1, L) q₄ (q₄, y, L) (q₅, y, R) (q₇, 1, L) (q₄, 1, L) q₅ (q₅, y, R) (q₃, y, L) (q₅, A, R) (q₅, 1, R) q₆ (q₆, y, R) (q₃, A, L) (q₆, A, R) (q₆, 1, R) q₇ (q₇, 0, R) (q₆, y, R) (q₇, 1, R) (q₂, 0, R) State set Q = {q₁, q₂, q₃, q₄, q₅, q₆, q₇}. Alphabet A = {y, 0, 1, A}.

TABLE 3 Boolean Universal Turing Machine program η = ( η ₀ η ₁ η ₂, η ₃ η ₄, η ₅) 10 00 01 11 001 (001, 00, 0) (001, 00, 0) (010, 01, 0) (001, 01, 0) 010 (001, 00, 0) (010, 10, 1) (010, 11, 1) (110, 10, 1) 011 (011, 10, 0) (000, 00, h) (011, 11, 0) (100, 01, 0) 100 (100, 10, 0) (101, 10, 1) (111, 01, 0) (100, 01, 0) 101 (101, 10, 1) (011, 10, 0) (101, 11, 1) (101, 01, 1) 110 (110, 10, 1) (011, 11, 0) (110, 11, 1) (110, 01, 1) 111 (111, 00, 1) (110, 10, 1) (111, 01, 1) (010, 00, 1) State set Q = {001, 010, 011, 100, 101, 110, 111 }. A = {10, 00, 01, 11}.

TABLE 4 AEM Separation Rules for Level Set η ₀ ⁻¹{1} Fire Pattern Element (U, A_(i)) (W, A_(i)) (X, A_(i)) (Y, A_(i)) (Z, A_(i)) θ_(A) _(i) 111 10 A₀ 2 2 2 2 −2 7 111 01 A₁ 2 2 2 −2 2 7 111 00 A₂ 2 2 2 −2 −2 5 110 11 A₃ 2 2 −2 2 2 7 110 10 A₄ 2 2 −2 2 −2 5 110 01 A₅ 2 2 −2 −2 2 5 101 11 A₆ 2 −2 2 2 2 7 101 10 A₇ 2 −2 2 2 −2 5 101 01 A₈ 2 −2 2 −2 2 5 100 11 A₉ 2 −2 −2 2 2 5 100 10 A₁₀ 2 −2 −2 2 −2 3 100 01 A₁₁ 2 −2 −2 −2 2 3 100 00 A₁₂ 2 −2 −2 −2 −2 1 011 11 A₁₃ −2 2 2 2 2 7 010 11 A₁₄ −2 2 −2 2 2 5

TABLE 5 AEM Separation Rules for Level Set η ₁ ⁻¹{1} Fire Pattern Element (U, B_(i)) (W, B_(i)) (X, B_(i)) (Y, B_(i)) (Z, B_(i)) θ_(B) _(i) 111 11 B₀ 2 2 2 2 2 9 111 10 B₁ 2 2 2 2 −2 7 111 01 B₂ 2 2 2 −2 2 7 111 00 B₃ 2 2 2 −2 −2 5 110 11 B₄ 2 2 −2 2 2 7 110 10 B₅ 2 2 −2 2 −2 5 110 01 B₆ 2 2 −2 −2 2 5 110 00 B₇ 2 2 −2 −2 −2 3 101 00 B₈ 2 −2 2 −2 −2 3 100 01 B₉ 2 −2 −2 −2 2 3 011 10 B₁₀ −2 2 2 2 −2 5 011 01 B₁₁ −2 2 2 −2 2 5 010 11 B₁₂ −2 2 −2 2 2 5 010 01 B₁₃ −2 2 −2 −2 2 3 010 00 B₁₄ −2 2 −2 −2 −2 1 001 01 B₁₅ −2 −2 2 −2 2 3

TABLE 6 AEM Separation Rules for Level Set η ₂ ⁻¹{1} Fire Pattern Element (U, C_(i)) (W, C_(i)) (X, C_(i)) (Y, C_(i)) (Z, C_(i)) θ_(C) _(i) 111 10 C₀ 2 2 2 2 −2 7 111 01 C₁ 2 2 2 −2 2 7 110 00 C₂ 2 2 −2 −2 −2 3 101 11 C₃ 2 −2 2 2 2 7 101 10 C₄ 2 −2 2 2 −2 5 101 01 C₅ 2 −2 2 −2 2 5 101 00 C₆ 2 −2 2 −2 −2 3 100 01 C₇ 2 −2 −2 −2 2 3 100 00 C₈ 2 −2 −2 −2 −2 1 011 10 C₉ −2 2 2 2 −2 5 011 01 C₁₀ −2 2 2 −2 2 5 010 10 C₁₁ −2 2 −2 2 −2 3 001 11 C₁₂ −2 −2 2 2 2 5 001 10 C₁₃ −2 −2 2 2 −2 3 001 00 C₁₄ −2 −2 2 −2 −2 1

TABLE 7 AEM Separation Rules for Level Set η ₃ ⁻¹{1} Fire Pattern Element (U, D_(i)) (W, D_(i)) (X, D_(i)) (Y, D_(i)) (Z, D_(i)) θ_(D) _(i) 111 00 D₀ 2 2 2 −2 −2 5 110 00 D₁ 2 2 −2 −2 −2 3 110 01 D₂ 2 2 −2 −2 2 5 110 10 D₃ 2 2 −2 2 −2 5 101 00 D₄ 2 −2 2 −2 −2 3 101 01 D₅ 2 −2 2 −2 2 5 101 10 D₆ 2 −2 2 2 −2 5 100 00 D₇ 2 −2 −2 −2 −2 1 100 10 D₈ 2 −2 −2 2 −2 3 011 01 D₉ −2 2 2 −2 2 5 011 10 D₁₀ −2 2 2 2 −2 5 010 00 D₁₁ −2 2 −2 −2 −2 1 010 01 D₁₂ −2 2 −2 −2 2 3 010 11 D₁₃ −2 2 −2 2 2 5

TABLE 8 AEM Separation Rules for Level Set η ₄ ⁻¹{1} Fire Pattern Element (U, E_(i)) (W, E_(i)) (X, E_(i)) (Y, E_(i)) (Z, E_(i)) θ_(E) _(i) 111 01 E₀ 2 2 2 −2 2 7 110 11 E₁ 2 2 −2 2 2 7 110 01 E₂ 2 2 −2 −2 2 5 110 00 E₃ 2 2 −2 −2 −2 3 101 11 E₄ 2 −2 2 2 2 7 101 01 E₅ 2 −2 2 −2 2 5 100 11 E₆ 2 −2 −2 2 2 5 100 01 E₇ 2 −2 −2 −2 2 3 011 11 E₈ −2 2 2 2 2 7 011 01 E₉ −2 2 2 −2 2 5 010 01 E₁₀ −2 2 −2 −2 2 3 001 11 E₁₁ −2 −2 2 2 2 5 001 01 E₁₂ −2 −2 2 −2 2 3

TABLE 9 AEM Separation Rules for Level Set η ₅ ⁻¹{1} Fire Pattern Element (U, F_(i)) (W, F_(i)) (X, F_(i)) (Y, F_(i)) (Z, F_(i)) θ_(F) _(i) 111 11 F₀ 2 2 2 2 2 9 111 10 F₁ 2 2 2 2 −2 7 111 01 F₂ 2 2 2 −2 2 7 111 00 F₃ 2 2 2 −2 −2 5 110 11 F₄ 2 2 −2 2 2 7 110 10 F₅ 2 2 −2 2 −2 5 110 01 F₆ 2 2 −2 −2 2 5 101 11 F₇ 2 −2 2 2 2 7 101 10 F₈ 2 −2 2 2 −2 5 101 01 F₉ 2 −2 2 −2 2 5 100 00 F₁₀ 2 −2 −2 −2 −2 1 010 11 F₁₁ −2 2 −2 2 2 5 010 01 F₁₂ −2 2 −2 −2 2 3 010 00 F₁₃ −2 2 −2 −2 −2 1

TABLE 10 Adding Four Bits: X₀, X₁, X₂, X₃ X₀ X₁ X₂ X₃ C₀ C₁ C₂ C₃ P₀ P₁ P₂ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1

TABLE 11 Adding Four Bits: X_(k) to C_(j) Amplitudes and C_(j) Thresholds Ele- ment (X_(k), C₀) (X_(k), C₁) (X_(k), C₂) (X_(k), C₃) θ_(C) ₀ θ_(C) ₁ θ_(C) ₂ θ_(C) ₃ X_(k) 2 2 2 2 1 3 5 7 k index refers to X₀, X₁, X₂, X₃

TABLE 12 Adding Four Bits: C_(j) to P_(i) Amplitudes Element (C₀, P₀) (C₁, P₀) (C₂, P₀) (C₃, P₀) θ_(P) ₀ P₀ 2 −2 2 −2 1 Element (C₀, P₁) (C₁, P₁) (C₂, P₁) (C₃, P₁) θ_(P) ₁ P₁ 0 2 2 −4 1 Element (C₀, P₂) (C₁, P₂) (C₂, P₂) (C₃, P₂) θ_(P) ₂ P₂ 2 2 2 2 7

TABLE 13 Four Bit Multiplication y₃ y₂ y₁ y₀ * z₃ z₂ z₁ z₀ y₃z₀ y₂z₀ y₁z₀ y₀z₀ y₃z₁ y₂z₁ y₁z₁ y₀z₁ y₃z₂ y₂z₂ y₁z₂ y₀z₂ y₃z₃ y₂z₃ y₁z₃ y₀z₃ e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀

TABLE 14 Element E₀ represents bit e₀. Amplitude and Threshold (Y₀, E₀) (Z₀, E₀) θ_(E) ₀ 2 2 3

TABLE 15 e₁ Bit. S₁₀  

 y₁z₀. S₀₁  

 y₀z₁. S₁₀ S₀₁ C₀₁ C₁₁ 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 1

TABLE 16 e₁ Bit. S_(jk) to C_(il) Amplitudes and C_(il) Thresholds (S₀₁, C₀₁) (S₀₁, C₁₁) (S₁₀, C₀₁) (S₁₀, C₁₁) θ_(C) ₀₁ θ_(C) ₁₁ 2 2 2 2 1 3

TABLE 17 Element E₁ represents bit e₁. Amplitude and Threshold (C₀₁, E₁) (C₁₁, E₁) θ_(E) ₁ 2 −2 1

TABLE 18 e₂ Bit. S₂₀ 

 y₂z₀. S₁₁ 

 y₁z₁. S₀₂ 

 y₀z₂. S₂₀ S₁₁ S₀₂ C₁₁ C₀₂ C₁₂ C₂₂ C₃₂ P₀₂ P₁₂ P₂₂ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1

TABLE 19 e₂ Bit. S_(jk), C₁₁ to C_(i2) Amplitudes and C_(i2) Thresholds (S_(jk), (S_(jk), (S_(jk), (S_(jk), C₀₂) C₁₂) C₂₂) C₃₂) θ_(C) ₀₂ θ_(C) ₁₂ θ_(C) ₂₂ θ_(C) ₃₂ 2 2 2 2 1 3 5 7

TABLE 20 C_(j2) to P_(i2) Amplitudes and P_(i2) Thresholds Element (C₀₂, P₀₂) (C₁₂, P₀₂) (C₂₂, P₀₂) (C₃₂, P₀₂) θ_(P) ₀₂ P₀₂ 2 −2 2 −2 1 Element (C₀₂, P₁₂) (C₁₂, P₁₂) (C₂₂, P₁₂) (C₃₂, P₁₂) θ_(P) ₁₂ P₁₂ 0 2 2 −4 1 Element (C₀₂, P₂₂) (C₁₂, P₂₂) (C₂₂, P₂₂) (C₃₂, P₂₂) θ_(P) ₂₂ P₂₂ 2 2 2 2 7

TABLE 21 Element E₂ represents bit e₂. Amplitude and Threshold (P₀₂, E₂) θ_(E) ₂ 2 1

TABLE 22 e₃ Bit. S_(jk) 

 y_(j)z_(k) where j + k = 3. S₃₀ S₂₁ S₁₂ S₀₃ P₁₂ C₀₃ C₁₃ C₂₃ C₃₃ C₄₃ P₀₃ P₁₃ P₂₃ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 1 1 0 0 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1

TABLE 23 S_(jk) (j + k = 3). S_(jk) to C_(i3) connections and C_(i3) thresholds. (S_(jk), C_(i3)) θ_(C) ₀₃ θ_(C) ₁₃ θ_(C) ₂₃ θ_(C) ₃₃ θ_(C) ₄₃ 2 1 3 5 7 9

TABLE 24 C_(j3) to P_(i3) Amplitudes and P_(i3) Thresholds (C₀₃, P₀₃) (C₁₃, P₀₃) (C₂₃, P₀₃) (C₃₃, P₀₃) (C₄₃, P₀₃) θ_(P) ₀₃ 2 −2 2 −2 2 1 (C₀₃, P₁₃) (C₁₃, P₁₃) (C₂₃, P₁₃) (C₃₃, P₁₃) (C₄₃, P₁₃) θ_(P) ₁₃ 0 2 2 −4 0 1 (C₀₃, P₂₃) (C₁₃, P₂₃) (C₂₃, P₂₃) (C₃₃, P₂₃) (C₄₃, P₂₃) θ_(P) ₂₃ 2 2 2 2 2 7

TABLE 25 Element E₃ represents bit e₃. S_(jk) (j + k = 3). (P₀₃, E₃) θ_(E) ₃ 2 1

TABLE 26 e₄ Bit. S₃₁ 

 y₃z₁. S₂₂ 

 y₂z₂. S₁₃ 

 y₁z₃. S₃₁ S₂₂ S₁₃ P₂₂ P₁₃ C₀₄ C₁₄ C₂₄ C₃₄ C₄₄ P₀₄ P₁₄ P₂₄ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 1 1 0 0 0 0 1 0 1 0 0 1 0 1 1 0 0 0 0 1 0 0 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1

TABLE 27 e₄ Bit. S_(jk) to C_(i4) Amplitudes and C_(i4) Thresholds (S_(jk), C_(i4)) (P₂₂, C_(i4)) (P₁₃, C_(i4)) θ_(C) ₀₄ θ_(C) ₁₄ θ_(C) ₂₄ θ_(C) ₃₄ θ_(C) ₄₄ 2 2 2 1 3 5 7 9

TABLE 28 C_(j4) to P_(i4) Amplitudes and P_(i4) Thresholds (C₀₄, P₀₄) (C₁₄, P₀₄) (C₂₄, P₀₄) (C₃₄, P₀₄) (C₄₄, P₀₄) θ_(P) ₀₄ 2 −2 2 −2 2 1 (C₀₄, P₁₄) (C₁₄, P₁₄) (C₂₄, P₁₄) (C₃₄, P₁₄) (C₄₄, P₁₄) θ_(P) ₁₄ 0 2 2 −4 0 1 (C₀₄, P₂₄) (C₁₄, P₂₄) (C₂₄, P₂₄) (C₃₄, P₂₄) (C₄₄, P₂₄) θ_(P) ₂₄ 2 2 2 2 2 7

TABLE 29 Element E₄ represents bit e₄. S_(jk) (j + k = 4). (P₀₄, E₄) θ_(E) ₄ 2 1

TABLE 30 e₅ Bit. S₃₂ 

 y₃z₂. S₂₃ 

 y₂z₃. S₃₂ S₂₃ P₁₄ P₂₃ C₀₅ C₁₅ C₂₅ C₃₅ P₀₅ P₁₅ P₂₅ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1

TABLE 31 e₅ Bit. S_(jk), P₁₄ and P₂₃ to C_(i5) Amplitudes and C_(i5) Thresholds (S₃₂, (S₂₃, (P₁₄, (P₂₃, C_(i5)) C_(i5)) C_(i5)) C_(i5)) θ_(C) ₀₅ θ_(C) ₁₅ θ_(C) ₂₅ θ_(C) ₃₅ 2 2 2 2 1 3 5 7

TABLE 32 C_(j5) to P_(i5) Amplitudes and P_(i5) Thresholds (C₀₅, P₀₅) (C₁₅, P₀₅) (C₂₅, P₀₅) (C₃₅, P₀₅) θ_(P) ₀₅ 2 −2 2 −2 1 (C₀₅, P₁₅) (C₁₅, P₁₅) (C₂₅, P₁₅) (C₃₅, P₁₅) θ_(P) ₁₅ 0 2 2 −4 1 (C₀₅, P₂₅) (C₁₅, P₂₅) (C₂₅, P₂₅) (C₃₅, P₂₅) θ_(P) ₂₅ 2 2 2 2 7

TABLE 33 Element E₅ represents bit e₅. S_(jk) (j + k = 5). (P₀₅, E₅) θ_(E) ₅ 2 1

TABLE 34 e₆ Bit. S₃₃ 

 y₃z₃ S₃₃ P₁₅ P₂₄ C₀₆ C₁₆ C₂₆ P₀₆ P₁₆ 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 1 1 0 1 1 1 0 0 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1

TABLE 35 e₆ Bit. S_(jk) to C_(i6) Amplitudes and C_(i6) Thresholds (S₃₃, C_(i6)) (P₁₅, C_(i6)) (P₂₄, C_(i6)) θ_(C) ₀₆ θ_(C) ₁₆ θ_(C) ₂₆ 2 2 2 1 3 5

TABLE 36 C_(j6) to P_(i6) Amplitudes and P_(i6) Thresholds (C₀₆, P₀₆) (C₁₆, P₀₆) (C₂₆, P₀₆) θ_(P) ₀₆ 2 −2 2 1 (C₀₆, P₁₆) (C₁₆, P₁₆) (C₂₆, P₁₆) θ_(P) ₁₆ 0 2 2 1

TABLE 37 Element E₆ represents bit e₆. Amplitude and Threshold (P₀₆, E₆) θ_(E) ₆ 2 1

TABLE 38 Four Bit Multiplication. 1110 * 0111 1 1 1 0 * 0 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 0 0 0 1 0 e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀ 14 * 7 = 98 in Base 10.

TABLE 39 Four Bit Multiplication. 1011 * 1001 1 0 1 1 * 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 1 1 e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀ 11 * 9 = 99 in Base 10.

TABLE 40 Four Bit Multiplication. 1111 * 1110 1 1 1 1 * 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 0 1 0 0 1 0 e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀ 15 * 14 = 210 in Base 10.

TABLE 41 Four Bit Multiplication. 1111 * 1111 1 1 1 1 * 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 e₇ e₆ e₅ e₄ e₃ e₂ e₁ e₀ 15 * 15 = 225 in Base 10. 

1. A method of executing a computer program wherein one or more program instructions executed at distinct instances are computed differently.
 2. The method of claim 1 wherein said distinct instances are different times.
 3. The method of claim 1 wherein said distinct instances are at different parts of a machine executing said computer program.
 4. The method of claim 1 wherein said one or more program instructions add numbers.
 5. The method of claim 1 wherein said one or more program instructions help multiply numbers.
 6. The method of claim 1 wherein randomness helps determine how said one or more program instructions are computed differently.
 7. The method of claim 6 wherein said randomness is quantum.
 8. The method of claim 7 wherein said quantum randomness is produced with photons or by a semiconductor material.
 9. The method of claim 1 wherein dynamic interpretations help differently compute said one or more program instructions.
 10. The method of claim 1 wherein said computer program is a digital computer program.
 11. The method of claim 1 wherein one or more said computer program instructions use time as a parameter.
 12. The method of claim 10 wherein said computer program is expressed in one or more of the following languages: C, JAVA, C++, Fortran, assembly language, Ruby, Forth, LISP, Haskell, RISC machine instructions, javascript, java virtual machine, Python, Turing machine.
 13. The method of claim 1 wherein said one or more program instructions are executed with a machine comprising a multiplicity of computing elements; and a multiplicity of couplings communicatively connecting the multiplicity of computing elements to one another, such that the couplings are capable of transmitting messages between the multiplicity of computing elements.
 14. The method of claim 13 wherein said couplings are expressed by one or more connection commands.
 15. The method of claim 1 wherein said one or more program instructions are executed with a system comprising a multiplicity of computing elements; and a multiplicity of couplings communicatively connecting the multiplicity of computing elements to one another, such that the couplings are capable of transmitting messages between the multiplicity of computing elements.
 16. The method of claim 2 wherein said different times are indicated by one or more commands.
 17. The method of claim 16 wherein said one or more commands include at least one meta command or a program command.
 18. The method of claim 16 wherein said one or more commands include at least one connection or element command.
 19. The method of claim 6 wherein said randomness is generated by a physical process.
 20. The method of claim 19 wherein said physical process occurs inside a physical barrier that contains an active element machine or inside a chip that executes an active element machine
 21. The method of claim 19 wherein said physical process comes from the environment.
 22. A machine that executes a computer program wherein at least one of the program's instructions executed at distinct instances is computed differently.
 23. The machine of claim 22 wherein said machine is comprised of a multiplicity of computing elements; and a multiplicity of couplings communicatively connecting the multiplicity of computing elements to one another, such that the couplings are capable of transmitting messages between the multiplicity of computing elements.
 24. The machine of claim 22 wherein said computer program is expressed in one or more of the following languages: C, JAVA, C++, Fortran, assembly language, Ruby, Forth, LISP, Haskell, RISC machine instructions, javascript, java virtual machine, Python, Turing machine.
 25. The machine of claim 22 wherein said distinct instances are different times.
 26. The machine of claim 22 wherein said distinct instances use different parts of the computing machine.
 27. The machine of claim 22 wherein said computer program is an active element machine program.
 28. The machine of claim 22 wherein randomness is used to help differently compute said program instruction(s).
 29. The machine of claim 22 wherein said program instruction(s) can be expressed as one or more Boolean functions.
 30. The machine of claim 22 wherein said machine has an unbounded prime directed edge complexity.
 31. The machine of claim 22 wherein said machine's execution has no recurrent configurations.
 32. A computing machine that uses edge pattern substitution to execute a computer program.
 33. The machine of claim 32 wherein said machine finds prime directed edges.
 34. The machine of claim 33 wherein said prime directed edges are derived from a computer program.
 35. The machine of claim 32 that compiles a computer program into a finite set of prime directed edges.
 36. A method of executing a computer program with edge pattern substitution.
 37. The method of claim 36 wherein said method finds prime directed edges.
 38. A method of computing a Boolean function using level sets.
 39. The method of claim 38 wherein an active element machine computes said Boolean function.
 40. The method of claim 39 wherein parameter values of said active element machine are chosen to separate said level sets.
 41. The method of claim 39 wherein said active element machine differently computes said Boolean function at distinct instances.
 42. The method of claim 41 wherein different parameter values are used to differently compute said Boolean function at distinct instances.
 43. A method of computing a Boolean function with an active element machine.
 44. The method of claim 43 wherein said Boolean function multiplies two numbers.
 45. The method of claim 43 wherein said Boolean function helps execute a program instruction in a computer program.
 46. The method of claim 43 wherein one or more members in the same level set of the Boolean function are expressed with a pattern established during a window of time.
 47. A method for solving a graph problem wherein time is used in the commands that express the computation of the problem.
 48. The method of claim 47 wherein said graph problem is a traveling salesman problem.
 49. The method of claim 47 where said problem involves computing one or more Ramsey numbers.
 50. The method of claim 47 wherein one or more connections are used to represent edges in said graph.
 51. The method of claim 47 wherein one or more elements are used to represent vertices in said graph.
 52. A method of using randomness to help execute computer program instructions.
 53. The method of claim 52 wherein said randomness is quantum.
 54. The method of claim 52 wherein said randomness helps determine how at least one said program instruction is computed differently at different instances.
 55. The method of claim 54 wherein said distinct instances are different times.
 56. The method of claim 54 wherein said distinct instances are at different parts of a machine executing said program instruction.
 57. The method of claim 52 wherein said randomness is produced by a physical process.
 58. The method of claim 54 wherein said program instruction is executed with an active element machine.
 59. The method of claim 58 wherein said randomness helps determine element or connection parameter values in said machine.
 60. The method of claim 54 wherein dynamic interpretations help differently compute one or more program instructions.
 61. The method of claim 54 wherein randomness helps determine a firing pattern during a window of time.
 62. The method of claim 61 wherein one or more members in the same level set are expressed with said firing pattern.
 63. A machine that is Turing incomputable.
 64. The machine of claim 63 wherein said machine executes a computer program.
 65. The machine of claim 64 wherein said machine uses quantum randomness to execute said computer program.
 66. The machine of claim 65 wherein said computing machine is an active element machine.
 67. The machine of claim 64 wherein Turing incomputability is used to conceal the execution of one or more program instructions of said computer program.
 68. The machine of claim 66 wherein said active element machine uses said quantum randomness to generate one or more interpretations.
 69. The machine of claim 68 wherein said one or more interpretations compute a program instruction.
 70. The machine of claim 68 wherein said one or more interpretations compute edge pattern substitution.
 71. The machine of claim 68 wherein said one or more interpretations compute one or more computer program instructions.
 72. The machine of claim 71 wherein said computer program instructions are expressed in one or more of the following languages: C, JAVA, C++, Fortran, assembly language, Ruby, Forth, LISP, Haskell, RISC machine instructions, javascript, java virtual machine, Python, Turing machine.
 73. The machine of claim 68 wherein one or more level sets help design said one or more interpretations.
 74. A method of executing a Turing incomputable procedure.
 75. The method of claim 74 wherein said method executes a computer program.
 76. The method of claim 75 wherein said method uses quantum randomness to execute said computer program.
 77. The method of claim 75 wherein said method uses an active element machine program.
 78. The method of claim 74 wherein Turing incomputability is used to conceal the execution of one or more program instructions of said computer program.
 79. The method of claim 77 wherein said active element machine program uses said quantum randomness to generate one or more interpretations.
 80. The method of claim 75 wherein said method uses one or more interpretations to compute a program instruction.
 81. The method of claim 80 wherein said one or more interpretations compute edge pattern substitution.
 82. The method of claim 75 wherein one or more interpretations compute one or more computer program instructions.
 83. The method of claim 82 wherein said computer program instructions are expressed in one or more of the following languages: C, JAVA, C++, Fortran, assembly language, Ruby, Forth, LISP, Haskell, RISC machine instructions, javascript, java virtual machine, Python, Turing machine.
 84. The method of claim 82 wherein one or more level sets help design said one or more interpretations. 